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I know the first principle,

$$f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$

However, I don't know what to do next. Help.

aki
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2 Answers2

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You do not want to suppose that $x=a,$ for then you have $$\frac{f(x)-f(a)}{x-a}=\frac00,$$ and what in the world does that even mean?

Rather, you want to assume that $x\ne a,$ and consider what happens to $\frac{f(x)-f(a)}{x-a}$ as $x$ tends toward $a.$

As a hint to help you get started, note that $$\sin^2(x)-\sin^2(a)=\bigl(\sin(x)+\sin(a)\bigr)\bigl(\sin(x)-\sin(a)\bigr),$$ and so $$\frac{f(x)-f(a)}{x-a}=\bigl(\sin(x)+\sin(a)\bigr)\cdot\frac{\sin(x)-\sin(a)}{x-a}$$ for all $x\ne a.$ What happens to the parenthetical expression as $x$ tends to $a$? What about the factor $\frac{\sin(x)-\sin(a)}{x-a}$ (which has the form of a difference quotient)?

Cameron Buie
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    You have a small typo. Notice you state $\frac{f(x) - f(x)}{x - a}$ instead of $\frac{f(x) - f(a)}{x - a}$. –  Nov 16 '14 at 17:05
  • but i supposed x=a in previous questions. – aki Nov 16 '14 at 17:09
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    @It'sMe: The derivative is just a limit, and we know that limits are not about the point itself; they are about what happens when we approach the point. – Sujaan Kunalan Nov 16 '14 at 17:11
  • @Chantry: Thanks! Fixed. – Cameron Buie Nov 16 '14 at 17:15
  • @It'sMe: If we are trying to find a limit such as $$\lim_{x\to a}g(x),$$ and we happen to know that $g(x)$ is defined and continuous at $x=a,$ then we can simply suppose that $x=a$ to get the answer. In this case, though, $$\frac{f(x)-f(a)}{x-a}$$ isn't even defined when $x=a,$ so we certainly can't assume that $x=a,$ here. – Cameron Buie Nov 16 '14 at 17:17
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Hint firts proof that: if $F(x)=\sin(x)\implies F'(a)=\cos(a)$, now if $f(x)=\sin^2(x)$, then $$f'(a)=\lim_{x\to a}\dfrac{\sin^2(x)-\sin^2(a)}{x-a}=\lim_{x\to a}\dfrac{[\sin(x)+\sin(a)][\sin(x)-\sin(a)]}{x-a}=\lim_{x\to a}[\sin(x)+\sin(a)]\cdot\lim_{x\to a}\dfrac{[\sin(x)-\sin(a)]}{x-a} =2\sin(a)\cdot F'(a)=2\sin(a)\cos(a)$$

AsdrubalBeltran
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