# How do you find the derivative of f(x) = cos(pi/2)x using the chain rule?

Jan 25, 2016

For the function formulated in the question, the derivative is

$f ' \left(x\right) = 0$.

However, the question might have meant this derivative instead:

$f ' \left(x\right) = - \frac{\pi}{2} \sin \left(\frac{\pi}{2} x\right)$

#### Explanation:

If you meant the function the way that you had typed it, then $\cos \left(\frac{\pi}{2}\right)$ is just a coefficient of $x$ and

$\cos \left(\frac{\pi}{2}\right) = 0$.

Thus,

$f \left(x\right) = \cos \left(\frac{\pi}{2}\right) \cdot x = 0 \cdot x = 0$

You don't need the chain rule for this one since the derivative of $f \left(x\right) = 0$ is $f ' \left(x\right) = 0$.

======================

You might have also meant

$f \left(x\right) = \cos \left(\frac{\pi}{2} x\right)$

In this case, you can apply the chain rule as follows:

$f \left(x\right) = \cos \left(\textcolor{b l u e}{u}\right) \text{ where } \textcolor{b l u e}{u} = \frac{\pi}{2} x$

The derivative of $f \left(x\right)$ is the derivative of $\cos u$ multiplied with the derivative of $u$:

$f ' \left(x\right) = \left[\cos u\right] ' \cdot u '$

It holds

$\left[\cos u\right] ' = - \sin u = - \sin \left(\frac{\pi}{2} x\right)$

$\left[u\right] ' = \left[\frac{\pi}{2} x\right] ' = \frac{\pi}{2}$

$f ' \left(x\right) = \left[\cos u\right] ' \cdot u ' = - \frac{\pi}{2} \sin \left(\frac{\pi}{2} x\right)$