# Finding Derivatives of Basic Functions

### Examples

##### Example 1

Find $$\displaystyle \frac d {dx}\left(8x -3\right)$$

Step 1

Use the rule for linear functions.

We know that the derivative of any linear function is just the slope of the line. Consequently,

$$\frac d {dx}\left(8x -3\right) = 8$$

##### Example 2

Suppose $$f(x) = \sin 3x - 4\cos 7x$$. Find $$f'\left(\frac \pi 2\right)$$

Step 1

Find the $$f'(x)$$ using the rules (not the definition).

\begin{align*} f'(x) & = \frac d {dx} \left(\sin 3x\right) - \frac d {dx}\left( 4\cos 7x\right) && \mbox{(Difference Rule)}\\[6pt] & = \frac d {dx} \left(\sin 3x\right) - 4\cdot\frac d {dx}\left( \cos 7x\right) && \mbox{(Coefficient Rule)}\\[6pt] & = 3\cos 3x - 4\cdot\frac d {dx}\left( \cos 7x\right) && \mbox{(Derivative of the Sine)}\\[6pt] & = 3\cos 3x - 4\cdot(-7\sin 7x) && \mbox{(Derivative of the Cosine)}\\[6pt] & = 3\cos 3x +28\sin 7x \end{align*}

Step 2

Evaluate the derivative at $$x = \pi / 2$$.

\begin{align*} f'\left(\frac \pi 2\right) & = 3\cos\left(3\cdot \frac \pi 2\right) +28\sin\left(7\cdot \frac \pi 2\right)\\[6pt] & = 3\cos\left(\frac{3\pi} 2\right) +28\sin\left(\frac{7\pi} 2\right)\\[6pt] & = 3(0) + 28(-1)\\[6pt] & = -28 \end{align*}

Answer

$$\displaystyle f'\left(\frac \pi 2\right) = -28$$ when $$f(x) = \sin 3x - 4\cos 7x$$.

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