Question

Determinar a derivada com x=0
F(x)=senh^-1 (tg (x))

Answer 1

Vamos usar as seguintes propriedades :

Derivada da função trigonométrica inversa do seno de uma função u

$\displaystyle [\text{Sen}^{-1}(\text u) ]'= \frac{\text u'}{\sqrt{1-\text u^2}}$

Derivada da tangente de uma função u :

$[\text {Tg(u)}]' = \text{Sec}^2(\text u).\text u'$

lembrando que :

$\displaystyle \text{Sec(u)} = \frac{1}{\text{Cos(u)}}$

Temos a seguinte função :

$\text{F(x)}= \text{Sen}^{-1}[\text{Tg(x)}]$

Derivando :

$\displaystyle \text{F'(x)}= \frac{[\text{Tg(x)}]'}{\sqrt{1-\text{Tg(x)}^2}}$

$\displaystyle \text{F'(x)}= \frac{[\text{Tg(x)}]'}{\sqrt{1-\text{Tg(x)}^2}}$

$\displaystyle \text{F'(x)}= \frac{\text{Sec}^2(\text x) }{\sqrt{1-\text{Tg(x)}^2}}$

Vamos trabalhar nesse denominador:

$\displaystyle \sqrt{1-\text{Tg(x)}^2} = \sqrt{1 - \frac{ \text {Sen}^2(\text x )}{ \text {Cos}^2( \text x) }} = \sqrt{\frac{ \text {Cos}^2( \text x) -\text {Sen}^2(\text x )}{ \text {Cos}^2( \text x) }}$

$\displaystyle \sqrt{1-\text{Tg(x)}^2} = \frac{\sqrt{\text{Cos}^2(\text x ) - \text{Sen}^2(\text x )}}{\text{Cos}(\text x )}$

Portanto :

$\displaystyle \text{F'(x)}= \frac{\text{Sec}^2(\text x) }{\sqrt{1-\text{Tg(x)}^2}} \to \text{F'(x)} =\frac{\displaystyle \frac{1}{\text{Cos}^2(\text x)}}{\displaystyle \frac{\sqrt{\text{Cos}^2(\text x) - \text{Sen}^2(\text x) }}{\text{Cos(x)}}}$

$\displaystyle \text{F'(x)} = \frac{1}{\text{Cos}^2(\text x) }.\frac{\text{Cos(x)}}{\sqrt{\text{Cos}^2(\text x) - \text{Sen}^2(\text x)}}$

$\displaystyle \text{F'(x)} =\frac{1}{\text{Cos(x)}.\sqrt{\text{Cos}^2(\text x) - \text{Sen}^2(\text x)}}$

Substituindo x = 0 :

$\displaystyle \text{F'(0)} =\frac{1}{\text{Cos(0)}.\sqrt{\text{Cos}^2(0) - \text{Sen}^2(0)}}$

$\displaystyle \text{F'(0)} =\frac{1}{1.\sqrt{1 -0}}$

Portanto :

$\huge\boxed{\text{F'(0)} = 1 }$