# How do you find the derivative of y= sin(xcosx)?

##### 1 Answer
May 24, 2015

There are two ways I can think of to do this. Substitution, or just thinking as you go.

1)
If you want to make this look nicer, you can write this as:

$y = \sin \left(x \cos x\right) = \sin \left(u\right)$
where $u = u \left(x\right) = x \cos x$.

$\frac{d}{\mathrm{du}} \left[\sin u \left(x\right)\right] = \cos u \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

Then,
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[x \cos x\right] = x \cdot \frac{d}{\mathrm{dx}} \left[\cos x\right] + \cos x \cdot \frac{d}{\mathrm{dx}} \left[x\right]$

$= - x \sin x + \cos x$

So, substituting that back in, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[\sin \left(x \cos x\right)\right] = \cos \left(x \cos x\right) \cdot \left(- x \sin x + \cos x\right)$

$= - \cos \left(x \cos x\right) \cdot \left(x \sin x - \cos x\right)$

2)

Or, you can just take the derivative and do the chain rule as you go.

$\frac{d}{\mathrm{dx}} \left[\sin \left(x \cos x\right)\right] = \cos \left(x \cos x\right) \cdot \left[x \cdot - \sin x + \cos x \cdot 1\right]$

$= - \cos \left(x \cos x\right) \cdot \left(x \sin x - \cos x\right)$