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.

1 Answer
Apr 30, 2018

(dk)/(da)=-4cos^3a*sin^6a+5sin^4a*cos^5adkda=4cos3asin6a+5sin4acos5a

Explanation:

  • Let g(a)=sin^5ag(a)=sin5a

g'(a)=5sin^4a*color(blue)(cosacolor(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