Question

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

Answer 1

`dy/dx=-8cot^3 2x * csc^2 2x`

Explanation:

We use the chain rule twice to find the derivative.

Method 1:

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

`y=f{g[h(x)]}`
`=> y'=f'{g[h(x)]} * g'[h(x)] * h'(x)`.

So, if
`y=cot^4 2x=(cot2x)^4`

then
`=>dy/dx=d/dx(cot 2x)^4`
`color(white)[=>dy/dx]=4(cot 2x)^3 * d/dx(cot 2x)`
`color(white)[=>dy/dx]=4cot^3 2x * (-csc^2 2x) * d/dx(2x)`
`color(white)[=>dy/dx]=-4cot^3 2x * csc^2 2x * 2`
`color(white)[=>dy/dx]=-8cot^3 2x * csc^2 2x`

Method 2:

We can also write the chain rule using Leibniz notation :

`dy/dx=dy/(du) * (du)/dx` but we need another term, so we extend:

`color(white)(dy/dx)=dy/(du) * (du)/(dv) * (dv)/(dx)`

Given
`y=cot^4 2x=(cot2x)^4`

Let `u=cot v` and `v=2x`.

Then `y=u^4`.

We have

`dy/dx=dy/(du) * (du)/(dv) * (dv)/(dx)`
`color(white)(dy/dx)=4u^3 * (-csc^2 v) * 2`

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

`color(white)(dy/dx)=4(cot v)^3 * (-csc^2 v) * 2`

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

`color(white)(dy/dx)=4(cot 2x)^3 * (-csc^2 2x) * 2`
`color(white)(dy/dx)=-8cot^3 2x * csc^2 2x`

Since `cot=cos/sin` and `csc=1/sin`, the trig functions can not be simplified any further.