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Question: Let $f(x) = \sin^2x$. Use the definition of a derivative to find the derivative.

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

I have forgotten the trick to solving this one.

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up vote 2 down vote accepted

Applying $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B),$

$\sin^2(x+h)-\sin^2x=\sin(2x+h)\sin h$

So, $$\frac{d(\sin^2x)}{dx}=\lim_{h\to 0}\frac{\sin(2x+h)\sin h}h=\lim_{h\to 0}\sin(2x+h)\cdot \lim_{h\to 0}\frac{\sin h}h =\sin 2x$$

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$$\dfrac{\sin^2(x+h) - \sin^2(x)}h = \left(\sin(x+h) + \sin(x) \right) \left(\dfrac{\sin(x+h) - \sin(x)}h \right)$$ Now recall that $\sin(x)$ is continuous and that $\displaystyle \lim_{h \to 0} \dfrac{\sin(x+h) - \sin(x)}h = \cos(x)$.

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I like how you split things up. As h goes to 0 I can see that you get 2sin(x) but you technically multiply this by the limit 0/0. Although it is equal to $cos(x)$ more work must be shown before I can just claim that. – Joakim Dec 9 '12 at 6:15

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