Question

Derivar f(x)= x arctg √x

Answer 1

Queremos a derivada da seguinte função:

$f(x) = x \cdot arctan( \sqrt{x} )$

Utilizaremos a regra do produto:

$f'(x) = x' \cdot arctan( \sqrt{x} ) + arctan'( \sqrt{x} ) \cdot x \\ \\ f'(x) = 1 \cdot arctan( \sqrt{x} ) + arctan'( \sqrt{x} ) \cdot x \\ \\ f'(x) = arctan( \sqrt{x} ) + arctan'( \sqrt{x} ) \cdot x$

Agora calcularemos a derivada de $arctan( \sqrt{x} )$, utilizando a regra da cadeia:

$\frac{df(u)}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$

$f = arctan(u) \, ; \, u = \sqrt{x}$

$arctan'( \sqrt{x} ) = arctan'(u) \cdot u' \\ \\ arctan'( \sqrt{x} ) = (\frac{1}{1+ u^2} ) \cdot (\sqrt{x} )' \\ \\ arctan'( \sqrt{x} ) = ( \frac{1}{1 + (\sqrt{x} )^2} ) \cdot (x^{ \frac{1}{2} })' \\ \\ arctan'( \sqrt{x} ) = ( \frac{1}{1 + x} ) \cdot \frac{1}{2} \cdot x^{-\frac{1}{2}} \\ \\ arctan'( \sqrt{x} ) = ( \frac{1}{1 + x} ) \cdot \frac{1}{2 \cdot \sqrt{x} } \\ \\ arctan'( \sqrt{x} ) = \frac{1}{ 2\sqrt{x} \cdot (1+x)}$

Logo temos:

$f'(x) = arctan( \sqrt{x} ) + arctan'(\sqrt{x} ) \cdot x \\ \\ f'(x) = arctan(\sqrt{x}) + \frac{1}{2 \sqrt{x} \cdot (1+x)} \cdot x \\ \\ f'(x) = arctan(\sqrt{x}) + \frac{x}{\sqrt{x} \cdot (2 + 2x)} \\ \\ f'(x) = arctan(\sqrt{x}) + \frac{\sqrt{x}}{2+2x}$