Question

How do you find the derivative of `y = cos^3(3x+1)`?

Answer 1

`(dy)/(dx)=-9sin(3x+1)cos^2(3x+1)`

Explanation:

`y=cos^3(3x+1)`

Applying chain rule: `(dy)/(dx)=(dy)/(du)*(du)/(dx)`

Let `u=cos(3x+1),` then `(du)/(dx)=-3sin(3x+1)`

`(dy)/(du)=d/(du)u^3=3u^2=3cos^2(3x+1)`

`:.(dy)/(dx)=3cos^2(3x+1)*-3sin(3x+1)`

`=-9sin(3x+1)cos^2(3x+1)`