# 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

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 $$2\sin(x)$$?

• $x$ isn’t $\sin x$. – Randall Nov 17 '18 at 14:26
• No. There's quite clearly a "$\sin$" in there. $\sin^2(x) = (\sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule. – user3482749 Nov 17 '18 at 14:27

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

• Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks! – Arcturus Nov 17 '18 at 14:33

Maybe a more convincing example. Following your logics,

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

But don't we have

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

No it is a case

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

$$\sin^2 x$$ 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.

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 $$f(x)=x^2$$.

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.