# y=cot^4 2x Find dy/dx?

Dec 14, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 8 {\cot}^{3} 2 x \cdot {\csc}^{2} 2 x$

#### Explanation:

We use the chain rule twice to find the derivative.

## Method 1:

Recall the chain rule: If $y$ can be written as $f \left[g \left(x\right)\right]$, then
$y ' = f ' \left[g \left(x\right)\right] \cdot g ' \left(x\right)$. If the input to $g$ is again a function, we can continue recursively:

$y = f \left\{g \left[h \left(x\right)\right]\right\}$
$\implies y ' = f ' \left\{g \left[h \left(x\right)\right]\right\} \cdot g ' \left[h \left(x\right)\right] \cdot h ' \left(x\right)$.

So, if
$y = {\cot}^{4} 2 x = {\left(\cot 2 x\right)}^{4}$

then
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {\left(\cot 2 x\right)}^{4}$
$\textcolor{w h i t e}{\implies \frac{\mathrm{dy}}{\mathrm{dx}}} = 4 {\left(\cot 2 x\right)}^{3} \cdot \frac{d}{\mathrm{dx}} \left(\cot 2 x\right)$
$\textcolor{w h i t e}{\implies \frac{\mathrm{dy}}{\mathrm{dx}}} = 4 {\cot}^{3} 2 x \cdot \left(- {\csc}^{2} 2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)$
$\textcolor{w h i t e}{\implies \frac{\mathrm{dy}}{\mathrm{dx}}} = - 4 {\cot}^{3} 2 x \cdot {\csc}^{2} 2 x \cdot 2$
$\textcolor{w h i t e}{\implies \frac{\mathrm{dy}}{\mathrm{dx}}} = - 8 {\cot}^{3} 2 x \cdot {\csc}^{2} 2 x$

## Method 2:

We can also write the chain rule using Leibniz notation :

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ but we need another term, so we extend:

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$

Given
$y = {\cot}^{4} 2 x = {\left(\cot 2 x\right)}^{4}$

Let $u = \cot v$ and $v = 2 x$.

Then $y = {u}^{4}$.

We have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}$
$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = 4 {u}^{3} \cdot \left(- {\csc}^{2} v\right) \cdot 2$

And since $u$ is a function of $v$, we substitute:

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = 4 {\left(\cot v\right)}^{3} \cdot \left(- {\csc}^{2} v\right) \cdot 2$

But $v$ is a function of $x$ too, so we substitute again:

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = 4 {\left(\cot 2 x\right)}^{3} \cdot \left(- {\csc}^{2} 2 x\right) \cdot 2$
$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - 8 {\cot}^{3} 2 x \cdot {\csc}^{2} 2 x$

Since $\cot = \frac{\cos}{\sin}$ and $\csc = \frac{1}{\sin}$, the trig functions can not be simplified any further.