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Derivative of Sin x Cos x using limit definition


I'm stuck on it

asked Jun 17 '19 at 17:17 by Bob

Can you give us a little more clarity on how you're stuck?

Jonny D edited Jun 18 '19 18:16

Your question doesn't show any attempts you made on this problem

Tommy P edited Jun 18 '19 21:07


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2

Hint: \\[12pt]
\cos(\alpha + \beta) \; = \; \cos\alpha cos\beta \; sin\alpha \sin\beta \\[12pt]
\lim\limits_{x \to \0} \frac{\sin x}{x} \; = \; 1\\[10pt]

asked Jun 18 '19 at 21:05 by alex

I never though about writing the cosine function.

Jonny D edited Jun 18 '19

0


f(x)=\cos x \\
f'(x) = \cos x (\lim\limits_{h \to \0} \frac{\cos h -1}{h}) \,-\, \sin x (\lim\limits_{h \to \0} \frac{\sin h}{h})\\
f'(x)= \cos x(0) -\,\sin x (1)\\
f'(x)= -\sin x

asked Jun 19 '19 at 16:40 by Lucy

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