Well, that's it. How do you calculate $\frac{d}{dx} {\sin5x}$ using the limit formula for derivatives?
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}h$$
I managed to get a lot of sines and cosines using trigonometric identities, but got stuck halfway. Any help would be appreciated.
$\endgroup$1 Answer
$\begingroup$\begin{align*} \lim_{h\to 0} \frac{\sin(5x+5h)-\sin 5x}{h} &= \lim_{h\to 0} \frac{\sin 5x\cos 5h + \cos 5x\sin 5h-\sin 5x}{h} \\ &= \lim_{h\to 0} \left(\sin 5x\frac{\cos 5h-1}{h} + \cos 5x\frac{\sin 5h}{h}\right) \\ &= 5\lim_{h\to 0} \left(\sin 5x\frac{\cos 5h-1}{5h} + \cos 5x\frac{\sin 5h}{5h}\right) \\ &= 5\lim_{k\to 0} \left(\sin 5x\frac{\cos k-1}{k} + \cos 5x\frac{\sin k}{k}\right) \\ &= 5\cos 5x \end{align*} assuming you already know the limits $\lim_{k\to 0} \frac{\cos k-1}{k} = 0$ and $\lim_{k\to 0} \frac{\sin k}{k} =1$.
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