How do I find, by the definition of a derivative, the derivative of tanx? $$f'(x)=\lim_{\Delta x \to 0}{f(x+\Delta x)-f(x)\over \Delta x}=\lim_{\Delta x \to 0}{\tan(x+\Delta x)-\tan(x)\over \Delta x}=\lim_{\Delta x \to 0}{{\sin(x+\Delta x)\over \cos(x+\Delta x)}-{\sin(x)\over \cos(x)}\over \Delta x}$$
I tried using identities but I always reached something like ${0\over \cos(x+ \Delta x)\cos(x)}$ which is no good… Thanks for any help!
P.S. I'm only at the beginning of the derivatives subject, so we need to find the derivative of $tanx$ using definition only.
Best Answer
Taking the alternative and equivalent definition of derivative and using a little trigonometry and known limits , we get for $\,\displaystyle{\tan'(y):=\lim_{x\to y} \frac{\tan x-\tan y}{x-y}}\,$:
$$\frac{\tan x-\tan y}{x-y}=\frac{\tan(x-y)(1+\tan x\tan y)}{x-y}=$$
$$=\frac{\sin(x-y)}{x-y}\frac{1+\tan x\tan y}{\cos(x-y)}\xrightarrow [x\to y]{}1\cdot\frac{1+\tan^2y}{1}=1+\tan^2 y=\frac{1}{\cos^2 y}=\sec^2y$$