Derivada del tangente tan x

Derivada del tangente tan x

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

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

Esta igualdad se ha demostrado en https://www.math-linux.com/mathematics/limits/article/proof-of-limit-of-sin-x-x-1-as-x-approaches-0

Concluímos que:

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