Question

Obtenha a derivada da função f definida por
f(x) = 2(cos(x)+2)^3 - x^2

Avalie a derivada f' obtida em x = π (isto é, f'(π)) e responda o valor correto de f'(π).

Answer 1

Desejamos calcular a derivada da função $\large\displaystyle\begin{gathered}\tt f(x)=2(\cos(x)+2)^3-x^2 \end{gathered}$ no ponto $\large\displaystyle\begin{gathered}\tt x=\pi\end{gathered}$.

Vamos então calcular a derivada da função f(x) e para isso iremos aplicar a regra da cadeia.

$\Large\displaystyle\begin{gathered} \tt f'(x)=2\cdot\frac{d}{dx} \left( (\cos(x)+2)^3\right) -\frac{d}{dx}\left( x^2\right) \end{gathered}$

$\Large\displaystyle\begin{gathered} \tt f'(x)=2\cdot\left(3\left(\cos(x)+2\right)^2\frac{d}{dx} \left( \cos(x)+2\right)\right) -2x \end{gathered}$

$\Large\displaystyle\begin{gathered} \tt f'(x)=-6\left(\cos(x)+2\right)^2(\sin(x))-2x \end{gathered}$

Sabendo então a derivada da função f(x), logo f'(π) é igual a:

$\Large\displaystyle\begin{gathered} \tt f'(\pi)=-6\left(\cos(\pi)+2\right)^2(\sin(\pi))-2\pi \end{gathered}$

$\Large\displaystyle\begin{gathered} \tt f'(\pi)=-6\left(-1+2\right)^2(0)-2\pi \end{gathered}$

$\Large\displaystyle\begin{gathered}\therefore\boxed{ \tt f'(\pi)=-2\pi }\end{gathered}$

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