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

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Answer 1 · 2015-08-10T03:39:50

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

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:

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

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