Vrftjkry

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

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.

$$(fcirc g)’(x) = f’(g(x))cdot g’x$$

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

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

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

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

So, the expression becomes

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