# Question #4b5a4

Jan 30, 2017

$\cos \left(\sqrt{x}\right) - \frac{x \sin \left(\sqrt{x}\right)}{2 \sqrt{x}}$

#### Explanation:

I'm pretty sure you're looking for the derivative, seeing as this is posted under "Differentiating Trigonometric Functions".

Let $f \left(x\right) = x \cos \left(\sqrt{x}\right)$. This is a product of the functions

$x$, and $\cos \left(\sqrt{x}\right)$. So, using the product rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(x\right) ' \cos \left(\sqrt{x}\right) + x \left[\cos \left(\sqrt{x}\right)\right] '$

$= \cos \left(\sqrt{x}\right) + x \left[\cos \left(\sqrt{x}\right)\right] '$.

Now we need to find the derivative of $\cos \left(\sqrt{x}\right)$:

The chain rule states that, if $y$ is a function of $u$, and $u$ is a function of $x$, then

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$.

Let $y = \cos \left(\sqrt{x}\right)$ and $u = \sqrt{x}$. Then $y = \cos u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin u \cdot u ' = - \sin u \cdot \frac{1}{2 \sqrt{x}} = - \frac{\sin \left(\sqrt{x}\right)}{2 \sqrt{x}}$

Therefore,

$\frac{\mathrm{df}}{\mathrm{dx}} = \cos \left(\sqrt{x}\right) - \frac{x \sin \left(\sqrt{x}\right)}{2 \sqrt{x}}$