How to Use the Definition of the Derivative

Example 1

Use the first version of the definition of the derivative to find $$f'(3)$$ for $$f(x) = 5x^2$$

Replace the $$x$$'s with 3's in the definition.

$$ \begin{align*} f'(3) & = \displaystyle\lim_{\Delta x \to 0} \frac{f(3+\Delta x) - f(3)}{\Delta x} \end{align*} $$

Note: '$$\Delta x$$' is considered a single symbol. So replacing the $$x$$'s with 3's does not change the $$\Delta x$$'s

Evaluate the functions in the definition.

$$ \begin{align*} f'(3) & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{f(3+\Delta x)} - \red{f(3)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{5(3+\Delta x)^2} - \red{5(3^2)}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{5\left(9+6\Delta x + (\Delta x)^2\right)} - \red{45}}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\blue{45+30\Delta x + 5(\Delta x)^2} - \red{45}}{\Delta x} \end{align*} $$

Simplify until the denominator no longer approaches 0 when $$\Delta x \to 0$$

$$ \begin{align*} f'(3) & = \displaystyle\lim_{\Delta x \to 0} \frac{45+30\Delta x + 5(\Delta x)^2 - 45}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{30\Delta x + 5(\Delta x)^2}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} \frac{\Delta x\left(30 + 5\Delta x\right)}{\Delta x}\\[6pt] & = \displaystyle\lim_{\Delta x \to 0} (30 + \Delta x) \end{align*} $$

Evaluate the simpler limit.

$$\displaystyle\lim_{\blue {\Delta x \to 0}} (30 + \blue{\Delta x})= 30 + \blue 0 = 30$$

$$f'(3) = 30$$ when $$f(x) = 5x^2$$

Example 2

Use the second version of the definition of the derivative to find $$\frac{df}{dx}$$ for the function $$f(x) = \sqrt x$$

Evaluate the functions in the definition.

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

Rationalize the numerator.

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

Simplify until the denominator no longer approaches 0 as $$h\to 0$$.

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

Evaluate this simpler limit.

$$ \displaystyle\lim_{\blue{h\to 0}} \frac{1}{\sqrt{x+\blue h}+\sqrt x} = \frac{1}{\sqrt{x+\blue 0}+\sqrt x} = \frac{1}{\sqrt x+\sqrt x} = \frac{1}{2\sqrt x} $$

$$\displaystyle \frac{df}{dx} = \frac 1 {2\sqrt x}$$ when $$f(x) = \sqrt x$$.

Example 3

Use the third version of the definition of the derivative to evaluate $$f'(5)$$ when $$f(x) = \frac 1 x$$

Setup the definition of the derivative using the appropriate value for $$a$$.

Note that $$a = 5$$ in this exercise, since we are evaluating $$f'(5)$$.

$$ \begin{align*} f'(5) & = \displaystyle\lim_{x\to 5} \frac{f(x) - f(5)}{x - 5} \end{align*} $$

Evaluate $$f(x)$$ and $$f(5)$$.

$$ f'(5) = \displaystyle\lim_{x\to 5} \frac{\blue{f(x)} - \red{f(5)}}{x - 5} = \displaystyle\lim_{x\to 5} \frac{\blue{\frac 1 x} - \red{\frac 1 5}}{x - 5} $$

Simplify until the denominator no longer approaches 0 as $$x$$ approaches 5.

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

Evaluate the simpler limit.

$$ f'(5) = \displaystyle\lim_{\blue{x\to 5}} \left(-\frac 1 {5\blue x}\right) = - \frac 1 {5\blue{(5)}} = -\frac 1 {25} $$

$$f'(5) = -\frac 1 {25}$$ when $$f(x) = \frac 1 x$$.

Example 4

Use the first variation of the definition of the derivative to find $$\frac{d}{dx}\left(\sin 2x\right)$$ when $$x$$ is in radians.

Evaluate the functions in the definition of the derivative.

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

Use the Sum of Angles identity for the sine function.

Recall that $$\sin(A+B) = \sin A\cos B + \sin B\cos A$$.

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

Rearrange the numerator so the terms containing $$\sin 2x$$ are together. Then, factor out the $$\sin 2x$$.

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

Separate into two limits.

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

Evaluate the first limit using the techniques from Indeterminate Limits---Cosine Forms.

$$ \begin{align*} \lim_{\Delta x \to 0} \frac{\sin 2x \left(\cos 2\Delta x - 1\right)}{\Delta x} & = \blue{\lim_{\Delta x \to 0} \frac{\cos 2\Delta x - 1}{\Delta x}} \cdot \sin 2x \\[6pt] & = \blue{(0)} \cdot \sin 2x \\[6pt] & = 0 \end{align*} $$

Our derivative calculations now look like this:

$$\frac{d}{dx}\left(\sin 2x\right) = 0 + \displaystyle\lim_{\Delta x \to 0}\frac{\sin 2\Delta x \cos 2x}{\Delta x}$$

Evaluate the remaining limit using the techniques from Indeterminate Limits---Sine Forms.

$$ \begin{align*} \lim_{\Delta x \to 0}\frac{\sin 2\Delta x \cos 2x}{\Delta x} & = \red{\lim_{\Delta x \to 0}\frac{\sin 2\Delta x}{\Delta x}}\cdot \cos 2x\\[6pt] & = \red{(2)}\cdot \cos 2x\\ & = 2\cos 2x \end{align*} $$

$$\displaystyle \frac{df}{dx} = 2\cos 2x$$ when $$f(x) = \sin 2x$$ and $$x$$ is in radians.