# How do you find the derivative of f(x)=pitan(pir^2-5r)?

May 1, 2017

$\left(2 \setminus {\pi}^{2} x - 5 \setminus \pi\right) {\sec}^{2} \left(\setminus \pi {x}^{2} - 5 x\right)$

#### Explanation:

Assuming you meant $f \left(x\right) = \setminus \pi \tan \left(\setminus \pi {x}^{2} - 5 x\right)$

$\setminus \pi$ is a constant so we can ignore that. Since the inside of tangent is a function we have to use the chain rule. The derivative of $\tan \left(x\right)$ is ${\sec}^{2} \left(x\right)$ so we do get the following.
$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus \pi {\sec}^{2} \left(\setminus \pi {x}^{2} - 5 x\right) \left(\setminus \frac{d}{\mathrm{dx}} g \left(x\right)\right)$

Then we multiply the derivative of everything inside the parentheses which is
$\setminus \frac{d}{\mathrm{dx}} g \left(x\right) = 2 \setminus \pi r - 5$

Combining them we get
$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus \pi {\sec}^{2} \left(\setminus \pi {x}^{2} - 5 x\right) \left(2 \setminus \pi x - 5\right) = \left(2 \setminus {\pi}^{2} x - 5 \setminus \pi\right) {\sec}^{2} \left(\setminus \pi {x}^{2} - 5 x\right)$

You can expand it, but it might not look as clean.