What is the derivative of #z=sin^3(theta)#?

1 Answer
Aug 10, 2015

Answer:

#dz/(d theta)=3sin^2(theta)cos(theta)#

Explanation:

This follows from the Chain Rule: #d/dx(f(g(x)))=f'(g(x))*g'(x)#

For the function #sin^{3}(theta)#, if we let #g(theta)=sin(theta)# and #f(theta)=theta^{3}#, then #sin^{3}(theta)=f(g(theta))#.

Since #f'(theta)=3theta^{2}# and #g'(theta)=cos(theta)#, we get:

#dz/(d theta)=f'(g(theta)) * g'(theta) = 3sin^{2}(theta) * cos(theta)#.