Question

Bằng định nghĩa, tính đạo hàm của hàm sô \(y = \tan x\) tại điểm x bất kì, \(x \ne \frac{\pi }{2} + k\pi \,\,\,(k \in \mathbb{Z})\)

Answer 1

\(\begin{array}{l}f'({x_0}) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f({x_0})}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\tan x - \tan {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\tan x - \tan {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{\sin x}}{{\cos x}} - \frac{{\sin {x_0}}}{{\cos {x_0}}}}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{\sin x\cos {x_0} - \sin {x_0}\cos x}}{{\cos x\cos {x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{1}{{\cos x\cos {x_0}}} = \frac{1}{{{{\cos }^2}{x_0}}}\\ \Rightarrow f'(x) = (\tan x)' = \frac{1}{{{{\cos }^2}x}} = 1 + {\tan ^2}x\end{array}\)