Why does the derivative of $\sin^2x$ need the chain rule? Isn't it just $2\sin x$ by the power rule?

Question

It's probably something obvious and I'm gonna slap myself in the face again, but

Why is the first derivative of $\sin^2(x)$ calculated via the chain rule? Isn't it just a standard $x^a$ (power rule) case, and therefore just $2\sin(x)$?

Answer 1

$\sin x\ne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=\sin x$.

Answer 2

No it is a case

$$g(x)=[f(x)]^2\implies g’(x)=2f(x)f’(x)$$

Answer 3

$\sin^2 x$ means $(\sin x)^2$. You’re asked to differentiate with respect to $x$, not $\sin x$, so the Chain Rule is required. So, this is a case of a composite function.

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

In this case, you have

$$\big(u^2\big)’ = 2u\cdot u’$$

where $u = \sin x$.

If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.

Answer 4

No! More general you got for the function $f(x)=x^a$ we get as derivative

$$f'(x)=(x^a)'=a(x)^{a-1}\cdot(x)'=a(x)^{a-1}\cdot1=ax^{a-1}$$

This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))\cdot g'(x)$. Therefore for $f(x)=\sin^2(x)$ we got

$$f'(x)=(\sin^2(x))'=2\sin(x)\cdot(\sin(x))'=2\sin(x)\cdot\cos(x)=2\sin(x)\cos(x)$$

where the inner function is given by $g(x)=\sin(x)$ and the outer function $f(x)=x^2$.

Answer 5

Maybe a more convincing example. Following your logics,

$$((x^3)^2)'=2x^3.$$

But don't we have

$$((x^3)^2)'=(x^6)'=6x^5\ ?!$$

Answer 6

The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $\sin^2(x)=\sin(x)\cdot\sin(x)$. Now apply the product rule.