Question

Encontre a equação da reta tangente à função​

Answer 1

$\Large\boxed{\begin{array}{l}\sf f(x)=\dfrac{x}{\sqrt{1-x^2}}\\\\\sf f'(x)=\dfrac{1\cdot(\sqrt{1-x^2})-x\cdot\dfrac{-\backslash\!\!\!2x}{\backslash\!\!\!2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}\\\\\sf f'(x)=\dfrac{\sqrt{1-x^2}+\dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf f'(0)=\dfrac{\sqrt{1-0^2}+\dfrac{0^2}{\sqrt{1-0^2}}}{1-0^2}\\\\\sf f'(0)=\dfrac{1}{1}=1\end{array}}$

$\Large\boxed{\begin{array}{l}\sf y=y_p+f'(p)(x-x_p)\\\sf y=0+1\cdot( x-0)\\\sf y=x\end{array}}$