# Calculating the derivative of $\sin^3(312x^2)$ [closed]

Can you explain in detail please how to find the derivative of this function? $$\sin^3(312x^2)$$

• Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ? Commented Nov 17, 2018 at 15:49
• What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Commented Nov 17, 2018 at 15:49
• What is the derivative of $\sin^3(x)$? Commented Nov 17, 2018 at 15:50
• How to ask a good question. Commented Nov 17, 2018 at 15:50
• Try the Chain rule Commented Nov 17, 2018 at 15:53

The derivative of $$f(g(x))$$ is (if it exists) $$f'(g(x))g'(x)\tag1$$ (application of chain rule).

Here $$g(x)=312x^2$$ (can you find $$g'(x)$$?).

And $$f(x)=\sin^3(x)$$.

Note that we can write $$f(x)=h(k(x))$$ where $$h(x)=x^3$$ and $$k(x)=\sin x$$ so that according to $$(1)$$: $$f'(x)=h'(k(x))k'(x)$$

This must be enough.

This is a repeated application of the chain rule:

$$\frac{d}{dx} \sin^3(312x^2) = 3 \sin^2(312x^2) \cdot \frac{d}{dx} (\sin(312x^2)= 3 \sin^2(312x^2) \cdot \cos(312x^2) \cdot \frac{d}{dx} 312x^2 =\\ 3 \sin^2(312x^2) \cdot \cos(312x^2) \cdot 624x$$

$$y = \sin^3(312x^2)$$

This is a composite function. Whenever you have a composite function, use the Chain Rule.

$$(f\circ g)’(x) = f’(g(x))\cdot g’x$$

Let $$f(x) = x^3$$ and $$g(x) = \sin(312x^2)$$.

$$\implies y’ = 3\sin^2(312x^2)\cdot [\sin (312x^2)]’$$

Here, you have to apply the same technique again to simplify the second part.

$$[\sin(312x^2)]’ = \sin’(312x^2)\cdot (312x^2)’ = \cos(312x^2)\cdot 64x = 624x\cos(312x^2)$$

So, the expression becomes

$$y’ = 3\sin^2(312x^2)\cdot 624x\cos(312x^2) = 1872x\sin^2(312x^2)\cos(312x^2)$$