Question

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

Answer 1

`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)`.