﻿ How to Use the Definition of the Derivative. Visual Explanation with color coded examples - 22 Practice Problems explained step by step with interactive problems, showing all work.

# How to Use the Definition of the Derivative: Practice Problems

Step 1

Substitute in $$-1$$ for $$x$$ in the derivative definition.

$$f'(-1) = \displaystyle\lim_{\Delta x \to 0} \frac{f(-1+\Delta x) - f(-1)}{\Delta x}$$

Step 2

Evaluate the functions in the derivative.

\begin{align*} f'(-1) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(-1+\Delta x)} - \red{f(-1)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{(-1+\Delta x)^2 + 3(-1+\Delta x)} - \red{\left[(-1)^2 + 3(-1)\right]}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{(-1+\Delta x)^2 + 3(-1+\Delta x)} + \red{2}}{\Delta x} \end{align*}

Step 3

Simplify until each term in the numerator has a $$\Delta x$$ in it.

\begin{align*} f'(-1) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{(-1+\Delta x)^2} + \red{3(-1+\Delta x)} + 2}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{1-2\Delta x + (\Delta x)^2} \red{-3+3\Delta x} +2}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{-2\Delta x + (\Delta x)^2 +3\Delta x}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\Delta x + (\Delta x)^2}{\Delta x} \end{align*}

Step 4

Factor $$\Delta x$$ out of the numerator and simplify.

$$f'(-1)= \displaystyle\lim_{\Delta x \to 0} \frac{\Delta x + (\Delta x)^2}{\Delta x} = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\Delta x}(1 + \Delta x)}{\blue{\Delta x}} = \displaystyle\lim_{\Delta x \to 0} (1 + \Delta x)$$

Step 5

Evaluate this simpler limit.

$$f'(-1)= \displaystyle\lim_{\blue{\Delta x \to 0}} (1 + \blue{\Delta x}) = 1 + \blue 0 = 1$$

$$f'(-1) = 1$$ when $$f(x) = x^2 + 3x$$

Step 1

Evaluate the functions in the definition of the derivative.

\begin{align*} \frac d {dx}\left(-x^3\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(x+\Delta x)} - \red{f(x)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{-(x+\Delta x)^3} - \red{(-x^3)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{-(x+\Delta x)^3} + \red{x^3}}{\Delta x} \end{align*}

Step 2

Expand and simplify the numerator until each term has a $$\Delta x$$ in it.

\begin{align*} \frac d {dx}\left(-x^3\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{-(x+\Delta x)^3} + x^3}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{-(x^3 + 3x^2\Delta x + 3x(\Delta x)^2+(\Delta x)^3)} + x^3}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\red{-x^3} - 3x^2\Delta x - 3x(\Delta x)^2-(\Delta x)^3 + \red{x^3}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{- 3x^2\Delta x - 3x(\Delta x)^2-(\Delta x)^3}{\Delta x} \end{align*}

Step 3

Factor a $$\Delta x$$ out of the numerator and continue to simplify.

\begin{align*} \frac d {dx}\left(-x^3\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{- 3x^2\Delta x - 3x(\Delta x)^2-(\Delta x)^3}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\Delta x}(- 3x^2 - 3x\Delta x-(\Delta x)^2)}{\blue{\Delta x}}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} (- 3x^2 - 3x\Delta x-(\Delta x)^2) \end{align*}

Step 4

Evaluate this simpler limit.

\begin{align*} \frac d {dx}\left(-x^3\right) & = \displaystyle\lim_{\blue{\Delta x \to 0}} (- 3x^2 - 3x\blue{\Delta x}-(\blue{\Delta x})^2)\\[6pt] & = - 3x^2 - 3x\blue{(0)}-(\blue{0})^2\\[6pt] & = -3x^2 \end{align*}

$$f'(x) = -3x^2$$ when $$f(x) = -x^3$$

Step 1

Evaluate the functions in the derivative definition.

$$\frac d {dx} \left(\sqrt{x+3}\right) = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(x+\Delta x)} - \red{f(x)}}{\Delta x} = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\sqrt{(x+\Delta x) + 3}} - \red{\sqrt{x+3}}}{\Delta x}$$

Step 2

Rationalize the Numerator.

\begin{align*} \frac d {dx} \left(\sqrt{x+3}\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\sqrt{x+\Delta x + 3} - \sqrt{x+3}}{\Delta x} \cdot \frac{\blue{\sqrt{x+\Delta x + 3} + \sqrt{x+3}}}{\blue{\sqrt{x+\Delta x + 3} + \sqrt{x+3}}}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{(\sqrt{x+\Delta x + 3})^2 - (\sqrt{x+3})^2}{\Delta x(\sqrt{x+\Delta x + 3} + \sqrt{x+3})}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{x+\Delta x + 3 - (x+3)}{\Delta x(\sqrt{x+\Delta x + 3} + \sqrt{x+3})} \end{align*}

Step 3

Simplify until the original $$\Delta x$$ in the denominator is gone.

\begin{align*} \frac d {dx} \left(\sqrt{x+3}\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue x+\Delta x + \red 3 - \blue x - \red 3}{\Delta x(\sqrt{x+\Delta x + 3} + \sqrt{x+3})}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\Delta x}}{\blue{\Delta x}(\sqrt{x+\Delta x + 3} + \sqrt{x+3})}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac 1 {\sqrt{x+\Delta x + 3} + \sqrt{x+3}} \end{align*}

Step 4

Evaluate the simpler limit, then simplify.

\begin{align*} \frac d {dx} \left(\sqrt{x+3}\right) & = \displaystyle\lim_{\blue{\Delta x \to 0}} \frac 1 {\sqrt{x+\blue{\Delta x} + 3} + \sqrt{x+3}}\\[6pt] & = \frac 1 {\sqrt{x+3} + \blue 0 + \sqrt{x+3}}\\[6pt] & = \frac 1 {2\sqrt{x+3}} \end{align*}

$$\displaystyle \frac d {dx} \left(\sqrt{x+3}\right) = \frac 1 {2\sqrt{x+3}}$$

Step 1

Substitute 12 in for $$x$$ in the definition of the derivative.

$$f'(12) = \displaystyle\lim_{\Delta x \to 0} \frac{f(12+\Delta x) - f(12)}{\Delta x}$$

Step 2

Evaluate the functions in the definition.

\begin{align*} f'(12) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(12+\Delta x)} - \red{f(12)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\sqrt{3(12+\Delta x)}} - \red{\sqrt{3(12)}}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\sqrt{36+3\Delta x}} - \red{\sqrt{36}}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\sqrt{36+3\Delta x}} - \red 6}{\Delta x} \end{align*}

Step 3

Rationalize the numerator.

\begin{align*} f'(12) & = \displaystyle\lim_{\Delta x \to 0} \frac{\sqrt{36+3\Delta x} - 6}{\Delta x} \cdot \frac{\blue{\sqrt{36+3\Delta x} + 6}}{\blue{\sqrt{36+3\Delta x} + 6}}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{(\sqrt{36+3\Delta x})^2 - 36}{\Delta x(\sqrt{36+3\Delta x} + 6)}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{36+3\Delta x - 36}{\Delta x(\sqrt{36+3\Delta x} + 6)} \end{align*}

Step 4

Simplify until the original $$\Delta x$$ in the denominator is gone.

\begin{align*} f'(12) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{36}+3\Delta x - \blue{36}}{\Delta x(\sqrt{36+3\Delta x} + 6)}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{3\red{\Delta x}}{\red{\Delta x}(\sqrt{36+3\Delta x} + 6)}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{3}{\sqrt{36+3\Delta x} + 6} \end{align*}

Step 5

Evaluate this simpler limit, then simplify the result.

\begin{align*} f'(12) & = \displaystyle\lim_{\blue{\Delta x \to 0}} \frac 3 {\sqrt{36+3\blue{\Delta x}} + 6}\\[6pt] & = \frac 3 {\sqrt{36+3\blue{(0)}} + 6}\\[6pt] & = \frac 3 {\sqrt{36} + 6}\\[6pt] & = \frac 3 {6 + 6}\\[6pt] & = \frac 3 {12}\\[6pt] & = \frac 1 4 \end{align*}

$$f'(12) = \frac 1 4$$ when $$f(x) = \sqrt{3x}$$

Step 1

Substitute 2 in for $$t$$ in the definition of the derivative.

$$f'(2) = \displaystyle\lim_{\Delta t \to 0} \frac{f(2+\Delta t) - f(2)}{\Delta t}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(2) & = \displaystyle\lim_{\Delta t \to 0} \frac{\blue{f(2+\Delta t)} - \red{f(2)}}{\Delta t}\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \frac{\blue{\frac 1 {(2+\Delta t) + 4}} - \red{\frac 1 {2 + 4}}}{\Delta t}\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \frac{\blue{\frac 1 {6+\Delta t}} - \red{\frac 1 6}}{\Delta t} \end{align*}

Step 3

Factor the $$\Delta t$$ out of the denominator.

\begin{align*} f'(2) & = \displaystyle\lim_{\Delta t \to 0} \frac{\frac 1 {6+\Delta t} - \frac 1 6}{\blue{\Delta t}}\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \left[\blue{\frac 1 {\Delta t}}\cdot \left(\frac 1 {6+\Delta t} - \frac 1 6\right)\right] \end{align*}

Step 4

Subtract the fractions and simplify.

\begin{align*} f'(2) & = \displaystyle\lim_{\Delta t \to 0} \left[\frac 1 {\Delta t}\cdot \left(\frac 1 {6+\Delta t} - \frac 1 6\right)\right]\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \left[\frac 1 {\Delta t}\cdot \left(\frac{\blue 6}{\blue 6(6+\Delta t)} - \frac{\red{6+\Delta t}}{6\red{(6 + \Delta t)}}\right)\right]\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \left(\frac 1 {\Delta t}\cdot \frac {\blue 6 - \red{(6+\Delta t)}}{6(6 + \Delta t)}\right)\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \left(\frac 1 {\Delta t}\cdot \frac {6 - 6 - \Delta t}{6(6 + \Delta t)}\right)\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} \left(\frac 1 {\blue{\Delta t}}\cdot \frac {-\blue{\Delta t}}{6(6 + \Delta t)}\right)\\[6pt] & = \displaystyle\lim_{\Delta t \to 0} -\frac 1 {6(6 + \Delta t)} \end{align*}

Step 5

Evaluate the simpler limit.

$$f'(2) = \displaystyle\lim_{\blue{\Delta t \to 0}} -\frac 1 {6(6 + \blue{\Delta t})} = -\frac 1 {6(6 + \blue 0)} = -\frac 1 {36}$$

$$\displaystyle f'(2) = -\frac 1 {36}$$ when $$\displaystyle f(t) = \frac 1 {t+4}$$

Step 1

Evaluate the functions in the definition of the derivative.

\begin{align*} \frac d {dx}\left(\frac 2 {5x}\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(x+\Delta x)} - \red{f(x)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\frac 2 {5(x+\Delta x)}} - \red{\frac 2 {5x}}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{\frac 2 {5x + 5\Delta x}} - \red{\frac 2 {5x}}}{\Delta x} \end{align*}

Step 2

Factor the $$\Delta x$$ out of the denominator, then subtract the fractions.

\begin{align*} \frac d {dx}\left(\frac 2 {5x}\right) & = \displaystyle\lim_{\Delta x \to 0} \frac{\frac 2 {5x + 5\Delta x} - \frac 2 {5x}}{\blue{\Delta x}}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \left[\blue{\frac 1 {\Delta x}}\cdot\left(\frac 2 {5x + 5\Delta x} - \frac 2 {5x}\right)\right]\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \left[\frac 1 {\Delta x}\cdot\left(\frac{2\blue{(5x)}}{\blue{5x}(5x + 5\Delta x)} - \frac{2\red{(5x+5\Delta x)}}{5x\red{(5x+5\Delta x)}}\right)\right]\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \left[\frac 1 {\Delta x}\cdot\frac{\blue{10x} - \red{(10x+10\Delta x)}}{5x(5x+5\Delta x)}\right]\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \left[\frac 1 {\Delta x}\cdot\frac{10x - 10x-10\Delta x}{5x(5x+5\Delta x)}\right]\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \left[\frac 1 {\blue{\Delta x}}\cdot\frac{-10\blue{\Delta x}}{5x(5x+5\Delta x)}\right]\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{-10}{5x(5x+5\Delta x)} \end{align*}

Step 3

Evaluate the simpler limit.

\begin{align*} \frac d {dx}\left(\frac 2 {5x}\right) & = \displaystyle\lim_{\blue{\Delta x \to 0}} \frac{-10}{5x(5x+5\blue{\Delta x})}\\[6pt] & = \frac{-10}{5x(5x+5\blue{(0)})}\\[6pt] & = -\frac{10}{5x(5x)}\\[6pt] & = -\frac{10}{25x^2}\\[6pt] & = -\frac{2}{5x^2} \end{align*}

$$\displaystyle \frac d {dx}\left(\frac 2 {5x}\right) = -\frac 2 {5x^2}$$

Step 1

Evaluate the functions in the definition of the derivative.

\begin{align*} \frac d {d\theta} \left(\cos \theta\right) & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\blue{f(\theta+\Delta \theta)} - \red{f(\theta)}}{\Delta \theta}\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\blue{\cos(\theta+\Delta \theta)} - \red{\cos \theta}}{\Delta \theta} \end{align*}

Step 2

Use the Sum of Angles Identity for the Cosine to expand the numerator.

\begin{align*} \frac d {d\theta} \left(\cos \theta\right) & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\cos(\blue\theta+\red{\Delta \theta}) - \cos \theta}{\Delta \theta}\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\cos\blue\theta \cos\red{\Delta \theta} - \sin\blue\theta\sin\red{\Delta\theta} - \cos \theta}{\Delta \theta} \end{align*}

Step 3

Rearrange the numerator so the terms involving $$\cos \theta$$ are together, then factor out the $$\cos \theta$$.

\begin{align*} \frac d {d\theta} \left(\cos \theta\right) & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\blue{\cos\theta} \cos\Delta \theta - \sin\theta\sin\Delta\theta - \blue{\cos \theta}}{\Delta \theta}\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\blue{\cos\theta} \cos\Delta \theta - \blue{\cos \theta} - \sin\theta\sin\Delta\theta}{\Delta \theta}\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\blue{\cos\theta}\left(\cos\Delta \theta - \blue 1\right) - \sin\theta\sin\Delta\theta}{\Delta \theta} \end{align*}

Step 4

Split into two fractions.

\begin{align*} \frac d {d\theta} \left(\cos \theta\right) & = \displaystyle\lim_{\Delta \theta \to 0} \frac{\cos\theta\left(\cos\Delta \theta - 1\right) - \sin\theta\sin\Delta\theta}{\Delta \theta}\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \left(\frac{\cos\theta\left(\cos\Delta \theta - 1\right)}{\Delta \theta} - \frac{\sin\theta\sin\Delta\theta}{\Delta \theta}\right) \end{align*}

Step 5

In each fraction isolate the $$\frac{\sin x} x$$ or $$\frac{\cos x - 1}{x}$$ forms. Then evaluate each limit.

\begin{align*} \frac d {d\theta} \left(\cos \theta\right) & = \displaystyle\lim_{\Delta \theta \to 0} \left(\frac{\cos\theta\blue{\left(\cos\Delta \theta - 1\right)}}{\blue{\Delta \theta}} - \frac{\sin\theta\red{\sin\Delta\theta}}{\red{\Delta \theta}}\right)\\[6pt] & = \displaystyle\lim_{\Delta \theta \to 0} \left(\frac{\cos \theta} 1\cdot \frac{\blue{\cos\Delta \theta - 1}}{\blue{\Delta \theta}} - \frac{\sin \theta} 1 \cdot \frac{\red{\sin\Delta\theta}}{\red{\Delta \theta}}\right)\\[6pt] & = \cos\theta\cdot \blue{\lim_{\Delta \theta \to 0} \frac{\cos\Delta \theta - 1}{\Delta \theta}} - \sin \theta \cdot \red{\lim_{\Delta \theta \to 0} \frac{\sin\Delta\theta}{\Delta \theta}}\\[6pt] & = \cos\theta\cdot \blue{(0)} - \sin \theta \cdot \red{(1)}\\[6pt] & = -\sin \theta \end{align*}

$$\displaystyle \frac d {d\theta} \left(\cos \theta\right) = -\sin \theta$$ when $$\theta$$ is in radians.

Step 1

Substitute 0 for $$x$$ in the derivative definition.

$$f'(0)= \displaystyle\lim_{\Delta x \to 0} \frac{f(0+\Delta x) - f(0)}{\Delta x}$$

Step 2

Evaluate the functions in the derivative definition.

\begin{align*} f'(0) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(0+\Delta x)} - \red{f(0)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{e^{2(0+\Delta x)}} - \red{e^{2(0)}}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{e^{2\Delta x}} - \red{e^0}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{e^{2\Delta x}} - \red 1}{\Delta x} \end{align*}

Step 3

Evaluate the limit using the techniques from the lesson on Indeterminate Limits---Exponential Forms.

\begin{align*} f'(0) & = \displaystyle\lim_{\Delta x \to 0} \blue{\frac 2 2}\cdot\frac{e^{2\Delta x} - 1}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue 2} 1\cdot\frac{e^{2\Delta x} - 1}{\blue 2\Delta x}\\[6pt] & = \blue 2\cdot \lim_{\Delta x \to 0} \frac{e^{2\Delta x} - 1}{\blue 2\Delta x}\\[6pt] & = 2 \cdot \red{\lim_{\Delta x \to 0} \frac{e^{2\Delta x} - 1}{2\Delta x}}\\[6pt] & = 2 \cdot \red{(1)}\\[6pt] & = 2 \end{align*}

$$f'(0) = 2$$ when $$f(x) = e^{2x}$$

Step 1

Evaluate the functions in the derivative definition and simplify.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{4(x + h) + 7} - \red{(4x+7)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{4x + 4h + 7} \red{-4x-7}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{4h}} h\\[6pt] & = \displaystyle\lim_{h\to 0} 4 \end{align*}

Step 2

Evaluate the limit.

$$\frac{df}{dx} = \displaystyle\lim_{h\to 0} 4 = 4$$

$$\displaystyle \frac d {dx} \left(4x + 7\right) = 4$$

Step 1

Evaluate the functions in the limit definition.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{(x + h)^2 + 6} - \red{(x^2 + 6)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{(x + h)^2 + 6} \red{\,-\,x^2 - 6}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{(x + h)^2} \red{ - x^2}} h \end{align*}

Step 2

Expand and then simplify the numerator.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\blue{(x + h)^2} \red{ - x^2}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{x^2 +2xh + h^2} \red{ - x^2}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{2xh + h^2}} h \end{align*}

Step 3

Factor out an $$h$$ and simplify.

$$\frac{df}{dx} = \displaystyle\lim_{h\to 0} \frac{2xh + h^2} h = \displaystyle\lim_{h\to 0} \frac{\blue h(2x + h)} {\blue h} = \displaystyle\lim_{h\to 0} (2x + h)$$

Step 4

Evaluate the limit.

$$\frac{df}{dx}= \displaystyle\lim_{\blue{h\to 0}} (2x + \blue h) = 2x + \blue 0 = 2x$$

$$\displaystyle \frac d {dx} \left(x^2 + 6\right) = 2x\$$

Step 1

Substitute 0 in for $$x$$ in the definition of the derivative.

$$f'(0) = \displaystyle\lim_{h\to 0} \frac{f(0 + h) - f(0)} h$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(0) & = \displaystyle\lim_{h\to 0} \frac{\blue{f(0 + h)} - \red{f(0)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{f(h)} - \red{f(0)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sqrt{2h+1}} - \red{\sqrt{2(0)+1}}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sqrt{2h+1}} - \red{\sqrt 1 }} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sqrt{2h+1}} - \red 1} h \end{align*}

Step 3

Rationalize the numerator and simplify.

\begin{align*} f'(0) & = \displaystyle\lim_{h\to 0} \frac{\sqrt{2h+1} - 1} h \cdot \blue{\frac{\sqrt{2h+1} + 1}{\sqrt{2h+1} + 1}}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{(\sqrt{2h+1})^2 - 1}{h(\sqrt{2h+1} + 1)}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{2h+1 - 1}{h(\sqrt{2h+1} + 1)}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{2\blue h}{\blue h(\sqrt{2h+1} + 1)}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac 2 {\sqrt{2h+1} + 1} \end{align*}

Step 4

Evaluate the limit.

\begin{align*} f'(0) & = \displaystyle\lim_{\blue{h\to 0}} \frac 2 {\sqrt{2\blue h+1} + 1}\\[6pt] & = \frac 2 {\sqrt{2\blue{(0)}+1} + 1}\\[6pt] & = \frac 2 {\sqrt 1 + 1}\\[6pt] & = \frac 2 {1+1}\\[6pt] & = \frac 2 2\\[6pt] & = 1 \end{align*}

$$f'(0) = 1$$ when $$f(x) = \sqrt{2x+1}$$

Step 1

Evaluate the functions in the definition of the derivative.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sqrt{1 - 5(x + h)}} - \red{\sqrt{1-5x}}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sqrt{1 - 5x -5h}} - \red{\sqrt{1-5x}}} h \end{align*}

Step 2

Rationalize the numerator and simplify.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\sqrt{1 - 5x -5h} - \sqrt{1-5x}} {\red h} \cdot \blue{\frac{\sqrt{1 - 5x -5h} + \sqrt{1-5x}}{\sqrt{1 - 5x -5h} + \sqrt{1-5x}}}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{(\sqrt{1 - 5x -5h})^2 - (\sqrt{1-5x})^2}{\red h\blue{(\sqrt{1 - 5x -5h} + \sqrt{1-5x})}}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{1 - 5x -5h - (1-5x)}{h(\sqrt{1 - 5x -5h} + \sqrt{1-5x})}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{1 - 5x -5h - 1+5x}{h(\sqrt{1 - 5x -5h} + \sqrt{1-5x})}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{-5\blue h}{\blue h(\sqrt{1 - 5x -5h} + \sqrt{1-5x})}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{-5}{\sqrt{1 - 5x -5h} + \sqrt{1-5x}} \end{align*}

Step 3

Evaluate the limit and simplify.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{\blue{h\to 0}} \frac{-5}{\sqrt{1 - 5x -5\blue h} + \sqrt{1-5x}}\\[6pt] & = \frac{-5}{\sqrt{1 - 5x -5\blue{(0)}} + \sqrt{1-5x}}\\[6pt] & = \frac{-5}{\sqrt{1 - 5x} + \sqrt{1-5x}}\\[6pt] & = \frac{-5}{2\sqrt{1 - 5x}} \end{align*}

$$\displaystyle \frac d {dx}\left(\sqrt{1-5x}\right) = \frac{-5}{2\sqrt{1 - 5x}}$$

Step 1

Evaluate the functions in the definition of the derivative.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\frac 4 {2(x + h)+3}} - \red{\frac 4 {2x+3}}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\frac 4 {2x + 2h+3}} - \red{\frac 4 {2x+3}}} h \end{align*}

Step 2

Factor out the $$h$$ from the denominator, then subtract the fractions. As always, simplify the results.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{h\to 0} \frac{\frac 4 {2x + 2h+3} - \frac 4 {2x+3}} {\blue h}\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\blue{\frac 1 h} \cdot \left(\frac 4 {2x + 2h+3} - \frac 4 {2x+3}\right)\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h \cdot \left(\frac{4\blue{(2x+3)}} {(2x + 2h+3)\blue{(2x+3)}} - \frac {4\red{(2x + 2h+3)}} {\red{(2x + 2h+3)}(2x+3)}\right)\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h \cdot \frac{\blue{8x+12} - \red{(8x + 8h+12)}} {(2x + 2h+3)(2x+3)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h \cdot \frac{8x+12 - 8x - 8h-12} {(2x + 2h+3)(2x+3)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 {\blue h} \cdot \frac{ - 8\blue h} {(2x + 2h+3)(2x+3)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{ - 8} {(2x + 2h+3)(2x+3)}\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{ - 8} {(2x + 2h+3)(2x+3)} \end{align*}

Step 3

Evaluate the limit and simplify.

\begin{align*} \frac{df}{dx} & = \displaystyle\lim_{\blue{h\to 0}} \frac{ - 8} {(2x + 2\blue h+3)(2x+3)}\\[6pt] & = \frac{ - 8} {(2x + 2\blue{(0)}+3)(2x+3)}\\[6pt] & = \frac{ - 8} {(2x+3)(2x+3)}\\[6pt] & = \frac{ - 8} {(2x+3)^2} \end{align*}

$$\displaystyle \frac d {dx}\left( \frac 4 {2x+3}\right) = \frac{ - 8} {(2x+3)^2}$$

Step 1

Substitute 3 in for $$x$$ in the derivative definition.

$$f'(3) = \displaystyle\lim_{h\to 0} \frac{f(3 + h) - f(3)} h$$

Step 2

Evaluate the functions in the derivative definition.

\begin{align*} f'(3) & = \displaystyle\lim_{h\to 0} \frac{\blue{f(3 + h)} - \red{f(3)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\frac 1 {3 + h} - \frac 1 {(3+h)+2}} - \red{\left(\frac 1 3 - \frac 1 {3+2}\right)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\frac 1 {3 + h} - \frac 1 {5+h}} - \red{\frac 2 {15}}} h \end{align*}

Step 3

Factor out the $$h$$ from the denominator, then subtract the fractions. As always, simplify the results.

\begin{align*} f'(3) & = \displaystyle\lim_{h\to 0} \frac{\frac 1 {3 + h} - \frac 1 {5+h} - \frac 2 {15}} {\blue h}\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\blue{\frac 1 h}\cdot \left(\frac 1 {3 + h} - \frac 1 {5+h} - \frac 2 {15}\right)\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \left(\frac{\blue{15(5+h)}}{\blue{15(5+h)}(3 + h)} - \frac{\red{15(3+h)}} {\red{15(3+h)}(5+h)} - \frac{2(3+h)(5+h)} {15(3+h)(5+h)}\right)\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \frac{\blue{15(5+h)} - \red{15(3+h)} - 2(3+h)(5+h)} {15(3+h)(5+h)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \frac{\blue{75+15h} - \red{45-15h} - 2(15 + 8h + h^2)} {15(3+h)(5+h)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \frac{30 - 30 - 16h -2h^2} {15(3+h)(5+h)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \frac{- 16h -2h^2} {15(3+h)(5+h)}\right] \end{align*}

Step 4

Factor an $$h$$ out of the numerator, and continue simplifying.

\begin{align*} f'(3) & = \displaystyle\lim_{h\to 0} \left[\frac 1 h\cdot \frac{- 16h -2h^2} {15(3+h)(5+h)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \left[\frac 1 {\blue h}\cdot \frac{\blue h\left(-16 -2h\right)} {15(3+h)(5+h)}\right]\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{-16 -2h} {15(3+h)(5+h)} \end{align*}

Step 5

Evaluate the limit, and simplify.

\begin{align*} f'(3) & = \displaystyle\lim_{\blue{h\to 0}} \frac{-16 -2\blue h} {15(3+\blue h)(5+\blue h)}\\[6pt] & = \frac{-16 -2\blue{(0)}} {15(3+\blue 0)(5+\blue 0)}\\[6pt] & = -\frac{16} {15(3)(5)}\\[6pt] & = -\frac{16} {225} \end{align*}

$$\displaystyle f'(3) = -\frac{16}{225}$$ when $$\displaystyle f(x) = \frac 1 x - \frac 1 {x+2}$$

Step 1

Substitute $$4$$ in for $$x$$ in the definition of the derivative.

$$f'(4) = \displaystyle\lim_{h\to 0} \frac{f(4 + h) - f(4)} h$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(4) & = \displaystyle\lim_{h\to 0} \frac{\blue{f(4 + h)} - \red{f(4)}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sin\left(\pi(4 + h)\right)} - \red{\sin 4\pi}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sin\left(4\pi + 4h\right)} - \red{\sin 4\pi}} h \end{align*}

Step 3

Expand the numerator using the Sum of Angles for the Sine.

\begin{align*} f'(4) & = \displaystyle\lim_{h\to 0} \frac{\sin\left(\blue{4\pi} + \red{4h}\right) - \sin 4\pi} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\sin\blue{4\pi}\cos\red{4h}+\sin\red{4h}\cos\blue{4\pi} - \sin 4\pi} h \end{align*}

Step 4

Group the terms containing a $$\sin 4\pi$$ and then factor out the sine.

\begin{align*} f'(4) & = \displaystyle\lim_{h\to 0} \frac{\blue{\sin4\pi}\cos 4h+\sin 4h\cos 4\pi - \blue{\sin4\pi}} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sin4\pi}\cos 4h- \blue{\sin4\pi}+\sin 4h\cos 4\pi} h\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\blue{\sin4\pi}\left(\cos 4h- 1\right)+\sin 4h\cos 4\pi} h \end{align*}

Step 5

Split into two limits. Factor out the term without the $$h$$.

\begin{align*} f'(4) & = \displaystyle\lim_{h\to 0} \frac{\sin4\pi\left(\cos 4h- 1\right)+\sin 4h\cos 4\pi} h\\[6pt] & = \displaystyle\lim_{h\to 0} \left(\frac{\sin4\pi\left(\cos 4h- 1\right)} h+\frac{\sin 4h\cos 4\pi} h\right)\\[6pt] & = \displaystyle\lim_{h\to 0} \frac{\sin4\pi\left(\cos 4h- 1\right)} h+\lim_{h\to 0} \frac{\sin 4h\cos 4\pi} h\\[6pt] & = \sin4\pi\cdot\lim_{h\to 0} \frac{\cos 4h- 1} h +\cos4\pi\cdot\lim_{h\to 0} \frac{\sin 4h} h \end{align*}

Step 6

Evaluate each limit.

\begin{align*} f'(4) & = \sin4\pi\cdot\blue{\lim_{h\to 0} \frac{\cos 4h- 1} h} +\cos4\pi\cdot\red{\lim_{h\to 0} \frac{\sin 4h} h}\\[6pt] & = \sin4\pi\cdot\blue{(0)} +\cos4\pi\cdot\red{\lim_{h\to 0} \frac 4 4\cdot \frac{\sin 4h} h}\\[6pt] & = \red 4 \cos4\pi\cdot\red{\lim_{h\to 0} \frac{\sin 4h} {4h}}\\[6pt] & = 4 \cos4\pi\cdot\red{(1)}\\[6pt] & = 4 \cos4\pi\\[6pt] & = 4 \end{align*}

$$f'(4) = 4$$ when $$f(x) = \sin(\pi x)$$ and $$x$$ is in radians.

Step 1

Replace $$x$$ with $$\frac\pi 6$$ in the definition of the derivative.

$$f'\left(\frac \pi 6\right) = \displaystyle\lim_{h\to 0} \frac{f\left(\frac \pi 6 + h\right) - f\left(\frac \pi 6\right)} h$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'\left(\frac \pi 6\right) &= \displaystyle\lim_{h\to 0} \frac{\blue{f\left(\frac \pi 6 + h\right)} - \red{f\left(\frac \pi 6\right)}} h\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\left(6\left(\frac \pi 6 + h\right)\right)} - \red{\cos\left(6\cdot\frac \pi 6\right)}} h\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos(\pi + 6h)} - \red{\cos\pi}} h \end{align*}

Step 3

Expand the numerator using the Sum of Angles for the Sine.

\begin{align*} f'\left(\frac \pi 6\right) &= \displaystyle\lim_{h\to 0} \frac{\cos(\blue{\pi} + \red{6h}) - \cos\pi} h\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\cos\blue{\pi}\cos\red{6h} - \sin\blue{\pi}\sin\red{6h} - \cos\pi} h \end{align*}

Step 4

Group the terms containing $$\cos \pi$$ and then factor out the cosine.

\begin{align*} f'\left(\frac \pi 6\right) &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\pi}\cos 6h - \sin \pi\sin 6h - \blue{\cos\pi}} h\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\pi}\cos 6h - \blue{\cos\pi} - \sin \pi\sin 6h} h\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\pi}\left(\cos 6h - 1\right) - \sin \pi\sin 6h} h \end{align*}

Step 5

Split into two limits and factor out the terms that don't contain an $$h$$.

\begin{align*} f'\left(\frac \pi 6\right) &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\pi}\left(\cos 6h - 1\right) - \red{\sin \pi}\sin 6h} h\\[6pt] &= \displaystyle\lim_{h\to 0} \left(\frac{\blue{\cos\pi}\left(\cos 6h - 1\right)} h - \frac{\red{\sin \pi}\sin 6h} h\right)\\[6pt] &= \displaystyle\lim_{h\to 0} \frac{\blue{\cos\pi}\left(\cos 6h - 1\right)} h - \lim_{h\to 0} \frac{\red{\sin \pi}\sin 6h} h\\[6pt] &= \blue{\cos\pi}\cdot\lim_{h\to 0} \frac{\cos 6h - 1} h - \red{\sin \pi}\cdot\lim_{h\to 0} \frac{\sin 6h} h \end{align*}

Step 6

Evaluate each limit.

\begin{align*} f'\left(\frac \pi 6\right) & = \cos\pi\cdot\blue{\lim_{h\to 0} \frac{\cos 6h - 1} h} - \sin \pi\cdot\red{\lim_{h\to 0} \frac{\sin 6h} h}\\[6pt] & = \cos\pi\cdot\blue{(0)} - \sin \pi\cdot\red{\lim_{h\to 0} \frac 6 6\cdot\frac{\sin 6h} h}\\[6pt] & = - \red 6 \sin \pi\cdot\red{\lim_{h\to 0}\frac{\sin 6h} {6h}}\\[6pt] & = - 6 \sin \pi\cdot\red{(1)}\\[6pt] & = - 6 \sin \pi\\[6pt] & = - 6 (0)\\[6pt] & = 0 \end{align*}

$$f'\left(\frac\pi 6\right) = 0$$ when $$f(x) = \cos(6x)$$ and $$x$$ is in radians.

Step 1

Replace $$a$$ with $$\frac 1 2$$ in the definition of the derivative

$$f'\left(\frac 1 2\right) = \displaystyle\lim_{x\to \frac 1 2} \frac{f(x) - f\left(\frac 1 2\right)}{x-\frac 1 2}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'\left(\frac 1 2\right) & = \displaystyle\lim_{x\to \frac 1 2} \frac{\blue{f(x)} - \red{f\left(\frac 1 2\right)}}{x-\frac 1 2}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{\blue{x^2 + 2x + 3} - \red{\left[\left(\frac 1 2\right)^2 + 2\left(\frac 1 2\right)+3\right]}}{x-\frac 1 2}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{\blue{x^2 + 2x + 3} - \red{\frac{17} 4}}{x-\frac 1 2}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{x^2 + 2x -\frac 5 4}{x-\frac 1 2} \end{align*}

Step 3

Factor the numerator and denominator and divide out any common factors. To facilitate this, we'll first multiply by $$\frac 4 4$$ to clear the fractions.

\begin{align*} f'\left(\frac 1 2\right) & = \displaystyle\lim_{x\to \frac 1 2} \frac{x^2 + 2x -\frac 5 4}{x-\frac 1 2}\cdot \blue{\frac 4 4}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{4x^2 + 8x -5}{4x- 2}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{\blue{(2x - 1)}(2x + 5)}{2\blue{(2x- 1)}}\\[6pt] & = \displaystyle\lim_{x\to \frac 1 2} \frac{2x + 5}{2} \end{align*}

Step 4

Evaluate the limit.

\begin{align*} f'\left(\frac 1 2\right) & = \displaystyle\lim_{\blue{x\to \frac 1 2}} \frac{2\blue x + 5}{2}\\[6pt] & = \frac{2\blue{\left(\frac 1 2\right)} + 5}{2}\\[6pt] & = \frac{6}{2}\\[6pt] & = 3 \end{align*}

$$f'\left(\frac 1 2\right) = 3$$ when $$f(x) = x^2 + 2x + 3$$.

Step 1

Substitute $$-4$$ for $$a$$ in the definition of the derivative.

$$f'(-4) = \displaystyle\lim_{x\to -4} \frac{f(x) - f(-4)}{x-(-4)} = \displaystyle\lim_{x\to -4} \frac{f(x) - f(-4)}{x+4}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(-4) & = \displaystyle\lim_{x\to -4} \frac{\blue{f(x)} - \red{f(-4)}}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{\blue{4x^3 -2} - \red{\left(4(-4)^3 -2\right)}}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{\blue{4x^3 -2} - \red{\left(4(-64) -2\right)}}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{\blue{4x^3 -2} - \red{\left(-258\right)}}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{4x^3 + 256}{x+4} \end{align*}

Step 3

Factor the numerator. Divide out any factors that are common with the denominator.

\begin{align*} f'(-4) & = \displaystyle\lim_{x\to -4} \frac{4x^3 + 256}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{4(x^3 + 64)}{x+4}\\[6pt] & = \displaystyle\lim_{x\to -4} \frac{4\blue{(x+4)}(x^2 - 4x + 16)}{\blue{x+4}}\\[6pt] & = \displaystyle\lim_{x\to -4} 4(x^2 - 4x + 16) \end{align*}

Step 4

Evaluate the limit.

\begin{align*} f'(-4) & = \displaystyle\lim_{\blue{x\to -4}} 4(\blue x^2 - 4\blue x + 16)\\[6pt] & = 4(\blue{(-4)}^2 - 4\blue{(-4)} + 16)\\[6pt] & = 4(48)\\[6pt] & = 192 \end{align*}

$$f'(-4) = 192$$ when $$f(x) = 4x^3 -2$$

Step 1

Substitute 1 for $$a$$ in the definition of the derivative.

$$f'(1) = \displaystyle\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(1) & = \displaystyle\lim_{x\to 1} \frac{\blue{f(x)} - \red{f(1)}}{x-1}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{\blue{\sqrt{9x-2}} - \red{\sqrt{9(1)-2}}}{x-1}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{\blue{\sqrt{9x-2}} - \red{\sqrt 7}}{x-1} \end{align*}

Step 3

Rationalize the numerator, and simplify.

\begin{align*} f'(1) & = \displaystyle\lim_{x\to 1} \frac{\sqrt{9x-2} - \sqrt 7}{x-1} \cdot \blue{\frac{\sqrt{9x-2} + \sqrt 7}{\sqrt{9x-2} + \sqrt 7}}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{(\sqrt{9x-2})^2 - (\sqrt 7)^2}{(x-1)(\sqrt{9x-2} + \sqrt 7)}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{9x-2 - 7}{(x-1)(\sqrt{9x-2} + \sqrt 7)}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{9x-9}{(x-1)(\sqrt{9x-2} + \sqrt 7)}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{9\blue{(x-1)}}{\blue{(x-1)}(\sqrt{9x-2} + \sqrt 7)}\\[6pt] & = \displaystyle\lim_{x\to 1} \frac{9}{\sqrt{9x-2} + \sqrt 7} \end{align*}

Step 4

Evaluate the limit and simplify.

\begin{align*} f'(1) & = \displaystyle\lim_{\blue{x\to 1}} \frac 9 {\sqrt{9\blue x-2} + \sqrt 7}\\[6pt] & = \frac 9 {\sqrt{9\blue{(1)}-2} + \sqrt 7}\\[6pt] & = \frac 9 {\sqrt{9-2} + \sqrt 7}\\[6pt] & = \frac 9 {\sqrt7 + \sqrt 7}\\[6pt] & = \frac 9 {2\sqrt 7} \end{align*}

If we rationalize the denominator, this becomes $$f'(1) = \frac{9\sqrt 7}{14}$$.

$$\displaystyle f'(1) = \frac 9 {2\sqrt 7} = \frac{9\sqrt 7}{14}$$ when $$f(x) = \sqrt{9x-2}$$

Step 1

Substitute 45 in for $$a$$ in the definition of the derivative.

$$f'(45) = \displaystyle\lim_{x\to 45} \frac{f(x) - f(45)}{x-45}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(45) & = \displaystyle\lim_{x\to 45} \frac{\blue{f(x)} - \red{f(45)}}{x-45}\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{\blue{\frac 1 {\sqrt{5x}}} - \red{\frac 1 {\sqrt{5(45)}}}}{x-45}\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{\blue{\frac 1 {\sqrt{5x}}} - \red{\frac 1 {\sqrt{225}}}}{x-45}\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{\blue{\frac 1 {\sqrt{5x}}} - \red{\frac 1 {15}}}{x-45} \end{align*}

Step 3

Rationalize the numerator.

\begin{align*} f'(45) & = \displaystyle\lim_{x\to 45} \frac{\frac 1 {\sqrt{5x}} - \frac 1 {15}}{x-45} \cdot \blue{\frac{\frac 1 {\sqrt{5x}} + \frac 1 {15}}{\frac 1 {\sqrt{5x}} + \frac 1 {15}}}\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{\left(\frac 1 {\sqrt{5x}}\right)^2 - \left(\frac 1 {15}\right)^2}{(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{\frac 1 {5x} - \frac 1 {225}}{(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)} \end{align*}

Step 4

Write the denominator as a separate fraction.

\begin{align*} f'(45) & = \displaystyle\lim_{x\to 45} \frac{\frac 1 {5x} - \frac 1 {225}}{(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\\[6pt] & = \displaystyle\lim_{x\to 45} \left[\frac 1 {(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\left(\frac 1 {5x} - \frac 1 {225}\right)\right] \end{align*}

Step 5

Subtract the fractions.

\begin{align*} f'(45) & = \displaystyle\lim_{x\to 45} \left[\frac 1 {(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\left(\frac 1 {5x} - \frac 1 {225}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 45} \left[\frac 1 {(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\left(\frac{\blue{225}}{5x\blue{(225)}} - \frac{\red{5x}}{225\red{(5x)}}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 45} \left[\frac 1 {(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\cdot\frac{\blue{225} - \red{5x}}{1125x}\right] \end{align*}

Step 6

Factor the numerator and divide out common factors.

\begin{align*} f'(45) & = \displaystyle\lim_{x\to 45} \left[\frac 1 {(x-45)\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\cdot\frac{225 - 5x}{1125x}\right]\\[6pt] & = \displaystyle\lim_{x\to 45} \left[\frac 1 {\blue{(x-45)}\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\cdot\frac{\red{-5}\blue{(x - 45)}}{\red{1125}x}\right]\\[6pt] & = \displaystyle\lim_{x\to 45} \left[\frac 1 {\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)}\cdot\frac{\red{-1}}{\red{225}x}\right]\\[6pt] & = \displaystyle\lim_{x\to 45} \frac{-1}{225x\left(\frac 1 {\sqrt{5x}} + \frac 1 {15}\right)} \end{align*}

Step 7

Evaluate the limit, and simplify the result.

\begin{align*} f'(45) & = \displaystyle\lim_{\blue{x\to 45}} \frac{-1}{225\blue x\left(\frac 1 {\sqrt{5\blue x}} + \frac 1 {15}\right)}\\[6pt] & = \frac{-1}{225\blue{(45)}\left(\frac 1 {\sqrt{5\blue{(45)}}} + \frac 1 {15}\right)}\\[6pt] & = \frac{-1}{225(45)\left(\frac 1 {\sqrt{225}} + \frac 1 {15}\right)}\\[6pt] & = \frac{-1}{225(45)\left(\frac 1 {15} + \frac 1 {15}\right)}\\[6pt] & = \frac{-1}{225(\red{45})\left(\frac 2 {\red{15}}\right)}\\[6pt] & = \frac{-1}{225(\red 3)(2)}\\[6pt] & = \frac{-1}{225(6)}\\[6pt] & = \frac{-1}{1350} \end{align*}

$$\displaystyle f'(45) = -\frac 1 {1350}$$ when $$\displaystyle f(x) = \frac 1 {\sqrt{5x}}$$

Step 1

Substitute 2 in for $$a$$ in the definition of the derivative.

$$f'(2) = \displaystyle\lim_{x\to 2} \frac{f(x) - f(2)}{x-2}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(2) & = \displaystyle\lim_{x\to 2} \frac{\blue{f(x)} - \red{f(2)}}{x-2}\\[6pt] & = \displaystyle\lim_{x\to 2} \frac{\blue{\frac 1 {6x+5}} - \red{\frac 1 {6(2)+5}}}{x-2}\\[6pt] & = \displaystyle\lim_{x\to 2} \frac{\blue{\frac 1 {6x+5}} - \red{\frac 1 {17}}}{x-2} \end{align*}

Step 3

Factor out the denominator, then subtract the fractions.

\begin{align*} f'(2) & = \displaystyle\lim_{x\to 2} \frac{\frac 1 {6x+5} - \frac 1 {17}}{\blue{x-2}}\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\blue{\frac 1 {x-2}}\left(\frac 1 {6x+5} - \frac 1 {17}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\frac 1 {x-2} \left(\frac{\blue{17}}{\blue{17}(6x+5)} - \frac{\red{6x+5}}{17\red{(6x+5)}}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\frac 1 {x-2} \cdot\frac{\blue{17} - \red{(6x+5)}}{17(6x+5)}\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\frac 1 {x-2} \cdot\frac{17 - 6x-5}{17(6x+5)}\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\frac 1 {x-2} \cdot\frac{12 - 6x}{17(6x+5)}\right] \end{align*}

Step 4

Factor the numerator then divide out the common factor.

\begin{align*} f'(2) & = \displaystyle\lim_{x\to 2} \left[\frac 1 {x-2} \cdot\frac{12 - 6x}{17(6x+5)}\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \left[\frac 1 {\blue{x-2}} \cdot\frac{-6\blue{(x-2)}}{17(6x+5)}\right]\\[6pt] & = \displaystyle\lim_{x\to 2} \frac{-6}{17(6x+5)} \end{align*}

Step 5

Evaluate the limit and simplify the result.

\begin{align*} f'(2) & = \displaystyle\lim_{\blue{x\to 2}} \frac{-6}{17(6\blue x+5)}\\[6pt] & = \frac{-6}{17(6\blue{(2)}+5)}\\[6pt] & = \frac{-6}{17(17)}\\[6pt] & = \frac{-6}{289} \end{align*}

$$\displaystyle f'(2) = -\frac 6 {289}$$ when $$\displaystyle f(x) = \frac 1 {6x + 5}$$

Step 1

Substitute 3 in for $$a$$ in the definition of the derivative.

$$f'(3) = \displaystyle\lim_{x\to 3} \frac{f(x) - f(3)}{x-3}$$

Step 2

Evaluate the functions in the definition of the derivative.

\begin{align*} f'(3) & = \displaystyle\lim_{x\to 3} \frac{\blue{f(x)} - \red{f(3)}}{x-3}\\[6pt] & = \displaystyle\lim_{x\to 3} \frac{\blue{\frac 3 {x^2+1}} - \red{\frac 3 {3^2 + 1}}}{x-3}\\[6pt] & = \displaystyle\lim_{x\to 3} \frac{\blue{\frac 3 {x^2+1}} - \red{\frac 3 {10}}}{x-3} \end{align*}

Step 3

Factor out the denominator, then subtract the fractions.

\begin{align*} f'(3) & = \displaystyle\lim_{x\to 3} \frac{\frac 3 {x^2+1} - \frac 3 {10}}{\blue{x-3}}\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\blue{\frac 1 {x-3}}\left(\frac 3 {x^2+1} - \frac 3 {10}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\left(\frac{3\blue{(10)}} {\blue{(10)}(x^2+1)} - \frac{3\red{(x^2+1)}} {10\red{(x^2+1)}}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\left(\frac{30} {\blue{(10)}(x^2+1)} - \frac{3x^2+3} {10\red{(x^2+1)}}\right)\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\cdot\frac{30 - (3x^2+3)} {10(x^2+1)}\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\cdot\frac{30 - 3x^2-3} {10(x^2+1)}\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\cdot\frac{27 - 3x^2} {10(x^2+1)}\right] \end{align*}

Step 4

Factor the numerator and divide out any common terms.

\begin{align*} f'(3) & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\cdot\frac{27 - 3x^2} {10(x^2+1)}\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {x-3}\cdot\frac{-3(x^2 - 9)} {10(x^2+1)}\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \left[\frac 1 {\blue{x-3}}\cdot\frac{-3(x+3)\blue{(x-3)}} {10(x^2+1)}\right]\\[6pt] & = \displaystyle\lim_{x\to 3} \frac{-3(x+3)} {10(x^2+1)} \end{align*}

Step 5

Evaluate the limit and simplify the result.

\begin{align*} f'(3) & = \displaystyle\lim_{\blue{x\to 3}} \frac{-3(\blue x+3)} {10(\blue x^2+1)}\\[6pt] & = \frac{-3(\blue 3+3)} {10(\blue{3}^2+1)}\\[6pt] & = \frac{-18} {10(10)}\\[6pt] & = \frac{-18} {100}\\[6pt] & = \frac{-9} {50} \end{align*}

$$\displaystyle f'(3) = -\frac 9 {50}$$ when $$\displaystyle f(x) = \frac 3 {x^2+1}$$