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I know that $\lim_{x\to 0} \dfrac{\sin x}{x} = 1$. But I'm stuck in using the definition of derivative to evaluate $(\sin^2 x)'$. Will appreciate any helpful input. Thank you.

uohzxela
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3 Answers3

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$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

Hence, \begin{align} (\sin^2 x)' &= \lim_{h\to 0}\frac{\sin^2(x+h)-\sin^2 (x)}{h} \\ &=\lim_{h\to 0}\frac{\sin(2x+h)\sin(h)}{h} \\ &=\lim_{h\to 0}\sin(2x+h) \cdot \lim_{h\to 0}\frac{\sin h}{h}=(\sin 2x) \cdot (1)=\sin2x. \end{align}

In the second step,I have used the fact $(\sin^2A-\sin^2B)=\sin(A+B).\sin(A-B)$.

Community
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Aang
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    oh my.. thank you so much for the comprehensive answer and the 'trick' you used! – uohzxela Sep 18 '12 at 14:46
  • Hmmn. As an aside, it's interesting that the result to this can be expressed as $\sin 2x$ as usually one would write $2 \sin x \cos x$ - I know using the Chain rule you would get this answer and I would usually follow this result. When I say the question posed my initial thought was to put the chain rule answer and it jarred slightly to see it as $\sin 2x$ even though this is absolutely correct. – Autolatry Sep 18 '12 at 15:04
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This is another way, which is quite similar but i think its worth showing. $$(\textrm{sin}^2(x^2))'=\lim_{t\rightarrow x}\frac{f(x)-f(t)}{x-t}=\lim_{t\rightarrow x}\frac{\sin^2(x)-\sin^2(t)}{x-t}=$$ $$\lim_{t\rightarrow x}\frac{\sin(x+t)\sin(x-t)}{x-t}=\lim_{t\rightarrow x}\ \sin(x+t)\cdot\lim_{t\rightarrow x}\frac{\sin(x-t)}{x-t}=$$ $$\lim_{t\rightarrow x}\ \sin(x+t)\cdot\lim_{u\rightarrow 0}\frac{\sin(u)}{u}= \lim_{t\rightarrow x}\ \sin(x+x)\cdot 1= \sin(2x)$$

Integral
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Implement the formula:

1) $[f^n(x)]'=nf^{n-1}(x)\cdot f'(x)$

2) $(\sin(x))'=\cos(x)$

3) $\sin(x)\cos(x)=\sin(2x)$

we have:

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

Madrit Zhaku
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