# How do you differentiate [sin[(3x+7)^2]^3]?

Apr 6, 2018

$\frac{\mathrm{ds} \in \left\{{\left[{\left(3 x + 7\right)}^{2}\right]}^{3}\right\}}{\mathrm{dx}} = 18 {\left(3 x + 7\right)}^{5} \cos \left[{\left(3 x + 7\right)}^{6}\right]$

#### Explanation:

The notation on this problem is a little ambiguous. I am going to assume that

$\sin {\left[{\left(3 x + 7\right)}^{2}\right]}^{3} = \sin \left\{{\left[{\left(3 x + 7\right)}^{2}\right]}^{3}\right\}$

and

$\sin {\left[{\left(3 x + 7\right)}^{2}\right]}^{3} \ne {\left\{\sin \left[{\left(3 x + 7\right)}^{2}\right]\right\}}^{3}$

First let's simplify noting that

$\sin \left\{{\left[{\left(3 x + 7\right)}^{2}\right]}^{3}\right\} = \sin \left[{\left(3 x + 7\right)}^{6}\right]$.

This is a job for the chain rule. In fact we apply it twice.

$\frac{\mathrm{df} \left(g \left(h \left(x\right)\right)\right)}{\mathrm{dx}} = {f}^{'} \left(g \left(h \left(x\right)\right)\right) \cdot {g}^{'} \left(h \left(x\right)\right) \cdot {h}^{'} \left(x\right)$

Here, $f \left(x\right) = \sin x$, $g \left(x\right) = {x}^{6}$, and $h \left(x\right) = 3 x + 7$. This means that ${f}^{'} \left(x\right) = \cos x$, ${g}^{'} \left(x\right) = 6 {x}^{5}$, and ${h}^{'} \left(x\right) = 3$.

So

$\frac{\mathrm{ds} \in \left[{\left(3 x + 7\right)}^{6}\right]}{\mathrm{dx}} = \cos \left[{\left(3 x + 7\right)}^{6}\right] \cdot 6 {\left(3 x + 7\right)}^{5} \cdot 3$

$= 18 {\left(3 x + 7\right)}^{5} \cos \left[{\left(3 x + 7\right)}^{6}\right]$.