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I thought that because this is true: $$ \sin^2x=(\sin x)^2,$$ I could differentiate the expression like this: $$ \frac{d}{dx}\sin^2x=2\cos x.$$ But I am supposed to get $$ \sin(2x) \quad \text{or}\quad 2\sin x \cos x.$$ Why am I wrong?

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    $\begingroup$ Chain rule. You didn't "chain". $\endgroup$ Jan 25, 2014 at 11:38
  • $\begingroup$ @theven we can use product rule as well $\endgroup$ Jan 25, 2014 at 12:17

7 Answers 7

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You can only use the rule: $ \frac{d}{dx} x^n = nx^{n-1} $ if $x$ is just $x$ and not a function. In the case for $\sin^2{x}$ then as you write it in the form $(\sin{x})^2$ you can see that we can't use the power rule because it's not just an $x$ term being raised to a constant power, it's a function being raised to a constant power. To differentiate $\sin^2{x}$ one must use the Chain Rule because $\sin{x}$ is a function of $x$ within another function (the function that is squaring $\sin{x}$) $x^2$. The chain rule is: $\frac{d}{dx}f(g(x))=g'(x)\times(f'(g(x))$ so you can apply that rule for this case with $f(x)=x^2$ and $g(x)=\sin{x}$.

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You are not using the chain rule correctly:

$$\frac{d}{dx}f(g(x))=\frac{df}{dx}(g(x))\frac{dg}{dx}(x)$$

Now $f(x)=x^2$ and $g(x)=\sin(x)$, therefore

$$\frac{d}{dx} (\sin(x))^2 = \overbrace{2\sin(x)}^{f^\prime(g(x))}\cdot\overbrace{\cos(x)}^{g^\prime(x)}$$

And to explain your other point of confusion: $\sin(2x)=2\sin(x)\cos(x)$ is a well-known trigonometric formula.

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The main way is using The Chain rule, but you may do it as follows :

$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ So $$(\sin^2(x))'=\left(\frac{1-\cos(2x)}{2}\right)'=\frac{1}2(1-\cos(2x))'=\frac{1}2(2\sin(2x))=\sin(2x)$$

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  • $\begingroup$ @SamiBenRomdhane: I'll say a pray for her at the midnight when all peoples are asleep. ان ناشئه الیل هی اشد وطئا... $\endgroup$
    – Mikasa
    Jan 25, 2014 at 17:58
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You are not applying properly the chain rule, in which $(f \circ g)'(x) = f'(g(x)) · g'(x)$, being $f'(x) = \frac {df(x)}{dx}$ and $g'(x) = \frac {dg(x)}{dx}$.

In particular, from $sin^{2}(x) = (x^2 \circ sin(x))$ it follows that $(sin^{2}(x))' = (x^2 \circ sin(x))' · sin'(x))$ $ = 2sin(x)cos(x) = sin(2x) $.

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You have to use chain rule. It is $\frac{d}{dx}(u(x))^n)=n(u(x))^{n-1} \frac{d}{dx}u(x)$

It gives $\frac{d}{dx}(\sin x)^2=2(\sin x)^{2-1} \frac{d}{dx}(\sin x)= 2\sin x \cos x$

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The noun is "derivative" and the verb is "differentiate".

I would advice never to write "$\sin^2(x)$" because it is in conflict with the usual notation $f^2(x) = f(f(x))$ for a generic function $f$. Also, if you write $\sin x$ remember that it means $\sin(x)$.

You need the chain rule because $\sin$ is a function. You want $\frac{d}{dx}(\sin(x)^2)$, which means you want to see how fast $\sin(x)^2$ changes with respect to $x$, so you can understand the chain rule to be first finding how fast $\sin(x)$ changes with respect to $x$, which is $\cos(x)$, and then multiply that by how fast $y^2$ changes with respect to $y$ where $y=\sin(x)$, which is $2y = 2 \sin(x)$, thus the product is $2 \sin(x) \cos(x)$.

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we can represent it as

$sin^2(x)=sin(x)*sin(x)$

now we know that $d(fg)=f'*g+g'*f$

so we have

$d(sin(x)*sin(x))=cos(x)*sin(x)+cos(x)*sin(x)=2sin(x)cos(x)=sin(2*x)$

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