Question

How do I find the derivative of k(a)=sin^5acos^4a?

How do I even find f'x for sin^5a and cos^4a? I would use the product rule, but I'm lost on just what the derivatives of the separate parts are.

Answer 1

`(dk)/(da)=-4cos^3a*sin^6a+5sin^4a*cos^5a`

Explanation:

• Let `g(a)=sin^5a`

`g'(a)=5sin^4a*color(blue)(cosa` `color(green)(rarr"Chain Rule")`

• Let `h(a)=cos^4a`

`h'(a)=4cos^3a*(color(blue)(-sina))` #color(green)(rarr"Chain Rule")#

`k=h(a)*g(a)`

`(dk)/(da)=h'(a)*g(a)+g'(a)*h(a)`

Substitute

`(dk)/(da)=(-4cos^3a*sina)*(sin^5a)+(5sin^4a*cosa)*(cos^4a)`

`(dk)/(da)=-4cos^3a*sin^6a+5sin^4a*cos^5a`