# Question #8f8da

Oct 18, 2017

$\pi \cdot {\sec}^{2} \left(\pi \cdot x\right)$

#### Explanation:

$\tan \left(\pi \cdot x\right) = \sin \frac{\pi \cdot x}{\cos} \left(\pi \cdot x\right)$

...so, you can use the rule for finding the derivative of the quotient of two functions.

if $f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$, then $f ' \left(x\right) = \frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{v {\left(x\right)}^{2}}$

so, for this function, you'd have:

$\frac{\pi \cdot \cos \left(\pi \cdot x\right) \cos \left(\pi \cdot x\right) + \sin \left(\pi \cdot x\right) \pi \sin \left(\pi \cdot x\right)}{\cos} ^ 2 \left(p \cdot x\right)$

Which simplifies to:

$\frac{\pi \left({\cos}^{2} \left(\pi \cdot x\right) + {\sin}^{2} \left(\pi \cdot x\right)\right)}{\cos} ^ 2 \left(\pi \cdot x\right)$

since ${\cos}^{2} \left(a\right) + {\sin}^{2} \left(a\right) = 1$, we can further simplify to:

$\frac{\pi}{\cos} ^ 2 \left(\pi \cdot x\right)$

and that is:

$\pi \cdot {\sec}^{2} \left(\pi \cdot x\right)$

GOOD LUCK