use definition of derivative to find a derivative of a trig function

**yvonnehr** · October 16th 2012, 05:31 PM

At this point, I can't use L'Hopital's rule. I tried to solve this using two different approaches and continue to get a 0 in the denominator even though I know that 0 should cancel out somehow. Here's the exercise:

Use the definition of the derivative to show that d/dx[cos x] = -sin x.
Hint: Use the limit laws in section 1.6 and the identity cos (A + B) = cos A cos B - sin A sin B.

Please help. This is one of two problems that I have left to do and have spent lots of time trying to solve to no avail.

**skeeter** · October 16th 2012, 05:50 PM

$f(x) = \cos{x}$ , $f(x+h) = \cos(x+h)$

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

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

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

$f'(x) = \lim_{h \to 0} \frac{\cos{x}(\cos{h} - 1) - \sin{x}\sin{h}}{h}$

$f'(x) = \lim_{h \to 0} \frac{\cos{x}(\cos{h} - 1)}{h} - \frac{\sin{x}\sin{h}}{h}$

finish it ...

**yvonnehr** · October 16th 2012, 11:00 PM

Sorry, but I don't know what to use to cancel out the h's in the denominators. I've gone through the entire section 1.6 and don't see anything that I believe I can use to address this. Should I be looking more closely at the trigonometric identities or properties to resolve this?

**skeeter** · October 17th 2012, 03:07 AM

you need two very important trig limits you've should have learned while covering limits ...

$\lim_{x \to 0} \frac{\sin{x}}{x} = 1$

$\lim_{x \to 0} \frac{1 - \cos{x}}{x} = 0$

**hollywood** · October 17th 2012, 06:51 AM

You can calculate these with L'Hôpital's rule, of course. That's what I do with $\frac{1-\cos{x}}{x}$ when I come across it. But I've seen $\frac{\sin{x}}{x}$ so many times that I just remember it.

- Hollywood

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