y=cot^4 2xy=cot42x Find dy/dxdydx?
1 Answer
Explanation:
We use the chain rule twice to find the derivative.
Method 1:
Recall the chain rule: If
So, if
then
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
Let
Then
We have
dy/dx=dy/(du) * (du)/(dv) * (dv)/(dx)
color(white)(dy/dx)=4u^3 * (-csc^2 v) * 2
And since
color(white)(dy/dx)=4(cot v)^3 * (-csc^2 v) * 2
But
color(white)(dy/dx)=4(cot 2x)^3 * (-csc^2 2x) * 2
color(white)(dy/dx)=-8cot^3 2x * csc^2 2x
Since