Derivative f’ of the function f(x)=tan x is: f’(x) = 1 + tan²x for any value x different of π/2 + kπ avec k ∈Z

Derivative of tan x

Derivative $f’$ of the function $f(x)=\tan x$ is: \(\forall x \neq \frac{\pi}{2}+k\pi, k \in \mathbb{Z}, f'(x) = 1+\tan ^{2} x\)


First we have: \((\tan x)' =\lim _{h \rightarrow 0} \dfrac{\tan (x+h) - \tan x }{h}\)

Now, let’s simplify the quotient:

\[\dfrac{\tan (x+h) - \tan x }{h}\]

Since $\tan (x+h)=\displaystyle\frac{\tan x+\tan h}{1-\tan x \tan h}$, we have: \(\begin{aligned} \frac{\tan (x+h) - \tan x }{h} &=\frac{\displaystyle\frac{\tan x+\tan h}{1-\tan x \tan h}-\tan x}{h} \\ &=\frac{\tan x+\tan h-\tan x(1-\tan x \tan h)}{h(1-\tan x \tan h)} \\ &=\frac{\tan x+\tan h-\tan x+\tan^{2} x \tan h}{h(1-\tan x \tan h)} \\ &=\frac{\tan h\left(1+\tan ^{2} x\right)}{h(1-\tan x \tan h)}\\ &=\frac{\displaystyle\frac{\sin h}{\cos h}\left(1+\tan ^{2} x\right)}{h(1-\tan x \tan h)}\\ &=\frac{\sin h \left(1+\tan ^{2} x\right)}{h \cos h(1-\tan x \tan h)} \\ &=\frac{\sin h}{h} \cdot \frac{1}{\cos h} \cdot \frac{1+\tan ^{2} x}{1-\tan x \tan h} \end{aligned}\)

Thus: \(\begin{aligned} (\tan x)' &=\lim _{h \rightarrow 0} \frac{\tan (x+h) - \tan x }{h}\\ &=\lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \frac{1}{\cos h} \cdot \frac{1+\tan ^{2} x}{1-\tan x \tan h}\\ &=\left(\lim _{h \rightarrow 0} \frac{\sin h}{h} \right) \cdot \left(\lim _{h \rightarrow 0} \frac{1}{\cos h} \right) \cdot \left(\lim _{h \rightarrow 0} \frac{1+\tan ^{2} x}{1-\tan x \tan h}\right)\\ &=\lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \frac{1}{\cos 0} \cdot \frac{1+\tan ^{2} x}{1-\tan x \tan 0}\\ &=1 \cdot 1 \cdot (1+\tan ^{2} x) \end{aligned}\)


\[\lim _{h \rightarrow 0} \frac{\sin h}{h}=1\]

This equality has been proved in /mathematics/limits/article/proof-of-limit-of-sin-x-x-1-as-x-approaches-0

We conclude: \((\tan x)' =\lim _{h \rightarrow 0} \dfrac{\tan (x+h) - \tan x }{h}=1+\tan ^{2} x\)