# How do you differentiate 3sin^2(3x)  using the chain rule?

May 25, 2018

$18 \sin \left(3 x\right) \cos \left(3 x\right)$

#### Explanation:

chain rule for 3 functions is $\frac{d}{\mathrm{dx}} \left(f \left(g \left(h \left(x\right)\right)\right)\right) = f ' \left(g \left(h \left(x\right)\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(h \left(x\right)\right)\right) = f ' \left(g \left(h \left(x\right)\right)\right) g ' \left(h \left(x\right)\right) h ' \left(x\right)$

in this problem, $f \left(x\right) = 3 {x}^{2}$, $g \left(x\right) = \sin \left(x\right)$, and $h \left(x\right) = 3 x$
that means: $f ' \left(x\right) = 6 x$, $g ' \left(x\right) = \cos \left(x\right)$, and $h \left(x\right) = 3$

so $\frac{d}{\mathrm{dx}} \left(3 {\sin}^{2} \left(3 x\right)\right) = 6 \left(\sin \left(3 x\right)\right) \cdot \cos \left(3 x\right) \cdot 3$
$= 18 \sin \left(3 x\right) \cos \left(3 x\right)$