# How do you differentiate y = (sin(3x) + cot(x^3))^8?

##### 1 Answer
Jul 31, 2015

${y}^{'} = 24 \cdot {\left[\sin \left(3 x\right) + \cot \left({x}^{3}\right)\right]}^{7} \cdot \left[\cos \left(3 x\right) - {x}^{2} \csc \left({x}^{3}\right)\right]$

#### Explanation:

Notice that you can write your function $y$ as

$y = {u}^{8}$, with $u = \sin \left(3 x\right) + \cot \left({x}^{3}\right)$

This means that you can use the power rule and the chain rule to differentiate $y$.

For a function $y$ that depends on a variable $u$, which in turn depends on a variable $x$, you can determine its derivative by using the chain rule

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(y\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)}$

In your case, this would be equivalent to

${y}^{'} = \frac{d}{\mathrm{du}} \left({u}^{8}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

${y}^{'} = 8 {u}^{7} \cdot \frac{d}{\mathrm{dx}} {\underbrace{\left(\sin \left(3 x\right) + \cot \left({x}^{3}\right)\right)}}_{\textcolor{red}{a \left(x\right)}}$

Now focus on differentiating function $\textcolor{red}{a \left(x\right)}$, which can be written as

$\frac{d}{\mathrm{dx}} \left(a\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(3 x\right)\right) + \frac{d}{\mathrm{dx}} \left(\cot \left({x}^{3}\right)\right)$

Once again, use the chain rule to differentiate these functions. Remember that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

and that

$\frac{d}{\mathrm{dx}} \left(\cot x\right) = - {\csc}^{2} x$

The first one will be

$\frac{d}{\mathrm{dx}} \left(\sin {u}_{1}\right) = \frac{d}{{\mathrm{du}}_{1}} \cdot \left(\sin {u}_{1}\right) \cdot \frac{d}{\mathrm{dx}} \left({u}_{1}\right)$, where ${u}_{1} = 3 x$

$\frac{d}{\mathrm{dx}} \left(\sin {u}_{1}\right) = \cos {u}_{1} \cdot \frac{d}{\mathrm{dx}} \left(3 x\right)$

$\frac{d}{\mathrm{dx}} \left(\sin \left(3 x\right)\right) = 3 \cos \left(3 x\right)$

The second one will be

$\frac{d}{\mathrm{dx}} \left(\cot {u}_{2}\right) = \frac{d}{{\mathrm{du}}_{2}} \left(\cot {u}_{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({u}_{2}\right)$, where ${u}_{2} = {x}^{3}$

$\frac{d}{\mathrm{dx}} \left(\cot {u}_{2}\right) = - {\csc}^{2} {u}_{2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$

$\frac{d}{\mathrm{dx}} \left(\cot {u}_{2}\right) = - {\csc}^{2} \left({x}^{3}\right) \cdot 3 {x}^{2}$

Plug these back into your tager derivative to get

${y}^{'} = 8 \cdot {\left[\sin \left(3 x\right) + \cot \left({x}^{3}\right)\right]}^{7} \cdot \left(3 \cos \left(3 x\right) - {\csc}^{2} \left({x}^{3}\right) \cdot 3 {x}^{2}\right)$

${y}^{'} = \textcolor{g r e e n}{24 \cdot {\left[\sin \left(3 x\right) + \cot \left({x}^{3}\right)\right]}^{7} \cdot \left[\cos \left(3 x\right) - {x}^{2} \csc \left({x}^{3}\right)\right]}$