La derivada f’ de la función f(x)=tan x es: f’(x) = 1 + tan²x para cualquier valor x real diferente de π / 2 + kπ, k ∈Z

La derivada $f’$ de la función $f(x)=\tan x$ es:

$\forall x \neq \frac{\pi}{2}+k\pi, k \in \mathbb{Z}, f'(x) = 1+\tan ^{2} x$

Prueba/Demostración

Primero tenemos:

$(\tan x)' =\lim _{h \rightarrow 0} \frac{\tan (x+h) - \tan x }{h}$

Ahora simplifiquemos el cociente:

$\frac{\tan (x+h) - \tan x }{h}$

Desde que tenemos $\tan (x+h)=\displaystyle\frac{\tan x+\tan h}{1-\tan x \tan h}$, entonces:

\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{\operatorname{sen} h}{\cos h}\left(1+\tan ^{2} x\right)}{h(1-\tan x \tan h)}\\ &=\frac{\operatorname{sen} h \left(1+\tan ^{2} x\right)}{h \cos h(1-\tan x \tan h)} \\ &=\frac{\operatorname{sen} h}{h} \cdot \frac{1}{\cos h} \cdot \frac{1+\tan ^{2} x}{1-\tan x \tan h} \end{aligned}

Por lo tanto:

\begin{aligned} (\tan x)' &=\lim _{h \rightarrow 0} \frac{\tan (x+h) - \tan x }{h}\\ &=\lim _{h \rightarrow 0} \frac{\operatorname{sen} h}{h} \cdot \frac{1}{\cos h} \cdot \frac{1+\tan ^{2} x}{1-\tan x \tan h}\\ &=\left(\lim _{h \rightarrow 0} \frac{\operatorname{sen} 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{\operatorname{sen} 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}

porque

$\lim _{h \rightarrow 0} \frac{\operatorname{sen} h}{h}=1$

$(\tan x)' =\lim _{h \rightarrow 0} \frac{\tan (x+h) - \tan x }{h}=1+\tan ^{2} x$