# How do you differentiate f(x)=sqrt(1-sin^2x)?

Apr 6, 2017

Apr 6, 2017

#### Explanation:

$f \left(x\right) = \sqrt{1 - {\sin}^{2} x} = \sqrt{{\cos}^{2} x} = \left\mid \cos x \right\mid = \left\{\begin{matrix}\cos x & \text{if" & cosx >=0 \\ -cosx & "if} & \cos x < 0\end{matrix}\right.$

So

$f \left(x\right) = \left\{\begin{matrix}\cos x & \text{if" & -pi/2+pik <= x <= pi/2+pik \\ -cosx & "if} & \frac{\pi}{2} + \pi k < x < \frac{3 \pi}{2} + \pi k\end{matrix}\right.$ where $k$ is an integer.

Therefore,

$f ' \left(x\right) = \left\{\begin{matrix}- \sin x & \text{if" & -pi/2+pik < x < pi/2+pik \\ sinx & "if} & \frac{\pi}{2} + \pi k < x < \frac{3 \pi}{2} + \pi k\end{matrix}\right.$ where $k$ is an integer.

$f$ is not differentiable at $x = \frac{\pi}{2} + \pi k$

Here is the graph of $f$

graph{abs(cosx) [-5.28, 5.82, -1.885, 3.664]}