y=cot^4 2xy=cot42x Find dy/dxdydx?

1 Answer
Dec 14, 2016

dy/dx=-8cot^3 2x * csc^2 2xdydx=8cot32xcsc22x

Explanation:

We use the chain rule twice to find the derivative.

Method 1:

Recall the chain rule: If yy can be written as f[g(x)]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.