# How do you differentiate y = 3x cos^2 (x)?

Feb 1, 2018

$3 \cos x$[$\cos x - 2 x \sin x$]

#### Explanation:

$d$[$u v$]= $v \mathrm{du} + u \mathrm{dv}$ this is the product rule, where v and u are both functions of x.

Let u =$3 x$ and $v = {\cos}^{2} x$

so we have ......${\cos}^{2} x$.[3]+$3 x$[$- 2 \sin x \cos x$]

=$3 {\cos}^{2} x$-$6 x \sin x \cos x$.....=3cosx[cosx-2xsinx].

${\cos}^{2} x$=${\left[\cos x\right]}^{2}$ and so need to use the chain rule to differentiate this, =$2 \cos x$, times the derivative of $\cos x$ which is -$\sin x$, so$\frac{d}{\mathrm{dx}} {\cos}^{2} x$$= - 2 \sin x \cos x$. hope this helped.