Question

Derive a função $f(x) = (x^2+1)^ \sqrt{x^2+5}$

Answer 1

Vamos reescrever a lei da função $f:$

$f(x)=(x^2+1)^{\sqrt{x^2+5}}\\\\ f(x)=\left(e^{\mathrm{\ell n}(x^2+1)} \right )^{\sqrt{x^2+5}}\\\\\\ f(x)=e^{\sqrt{x^2+5}\,\cdot\,\mathrm{\ell n}(x^2+1)}$


A função acima é uma função composta. Podemos enxergar assim:

$\left\{ \begin{array}{l} f(x)=e^{g(x)}\\\\ g(x)=\sqrt{x^2+5}\,\cdot\,\mathrm{\ell n}(x^2+1) \end{array} \right.$


Portanto, basta derivar $f$ usando a Regra da Cadeia:

$f'(x)=(e^{g(x)})'\\\\ f'(x)=e^{g(x)}\cdot g'(x)\\\\ f'(x)=f(x)\cdot g'(x)\\\\ f'(x)=(x^2+1)^{\sqrt{x^2+5}}\cdot \left[\sqrt{x^2+5}\cdot\mathrm{\ell n}(x^2+1) \right ]'$


Derivando $g$ pela Regra do Produto:

$f'(x)=(x^2+1)^{\sqrt{x^2+5}}\cdot \left[(\sqrt{x^2+5})'\cdot\mathrm{\ell n}(x^2+1)+\sqrt{x^2+5}\cdot (\mathrm{\ell n}(x^2+1))' \right ]\\\\\\ =(x^2+1)^{\sqrt{x^2+5}}\cdot \left[\left(\dfrac{1}{2\sqrt{x^2+5}}\cdot (x^2+5)' \right )\mathrm{\ell n}(x^2+1)+\sqrt{x^2+5}\left(\dfrac{1}{x^2+1}\cdot (x^2+1)' \right ) \right ]\\\\\\ =(x^2+1)^{\sqrt{x^2+5}}\cdot \left[\left(\dfrac{1}{2\sqrt{x^2+5}}\cdot 2x \right )\mathrm{\ell n}(x^2+1)+\sqrt{x^2+5}\left(\dfrac{1}{x^2+1}\cdot 2x \right ) \right ]\\\\\\ \therefore~~\boxed{\begin{array}{c}f'(x)=(x^2+1)^{\sqrt{x^2+5}}\cdot \left[\dfrac{x}{\sqrt{x^2+5}}\cdot\mathrm{\ell n}(x^2+1)+\sqrt{x^2+5}\cdot \dfrac{2x}{x^2+1} \right ] \end{array}}$