Why does the derivative of $\sin^2x$ need the chain rule? Isn't it just $2\sin x$ by the power rule?
It's probably something obvious and I'm gonna slap myself in the face again, but
$\endgroup$ 2Why 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 $2\sin(x)$?
6 Answers
$\begingroup$$\sin x\ne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=\sin x$.
$\endgroup$ 1 $\begingroup$Maybe a more convincing example. Following your logics,
$$((x^3)^2)'=2x^3.$$
But don't we have
$$((x^3)^2)'=(x^6)'=6x^5\ ?!$$
$\endgroup$ $\begingroup$No it is a case
$$g(x)=[f(x)]^2\implies g’(x)=2f(x)f’(x)$$
$\endgroup$ $\begingroup$$\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.
$\endgroup$ $\begingroup$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 $f(x)=x^2$.
$\endgroup$ $\begingroup$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.
$\endgroup$
