Question

Considerando a função f(x)=1/√x + √x o valor numérico de f'(1) é igual a

Answer 1

• O valor numérico de f'(1) é igual a 0.

Desejamos calcular a derivada da função: $\large\displaystyle\text{ \begin{gathered} f(x)=\frac{1}{\sqrt{x} }+\sqrt{x} \end{gathered} }$, no ponto f'(1).

Para calcular essa derivada, devemos lembrar algumas propriedades, as que iremos utilizar são:

             $\large\displaystyle\text{ \begin{gathered}\sqrt[n]{x^m} =x^{\frac{m}{n}} \end{gathered} }$

             $\large\displaystyle\begin{gathered}a^{-m}=\frac{1}{a^m} \end{gathered}$

Também é de suma importancia lembrar que a derivada da soma é a soma das derivadas. Logo:

                 $\large\displaystyle\text{ \begin{gathered} f'(x)=\left( \frac{1}{\sqrt{x}} \right)' +\left( \sqrt{x} \right)' \end{gathered} }$

• Aplicando as propriedades que eu havia dito, temos:

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( \frac{1}{\sqrt{x}} \right)' +\left( \sqrt{x} \right)' \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( \frac{1}{x^{\frac{1}{2}}} \right)' +\left( x^{\frac{1}{2}} \right)' \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( x^{-\frac{1}{2}} \right)' +\left( x^{\frac{1}{2}} \right)' \end{gathered} }$

Devemos agora aplicar a propriedade de derivação do tombo. Dada por:

               $\large\displaystyle\begin{gathered}f(x) =x^n\Leftrightarrow f'(x)= n\cdot x^{n - 1} \end{gathered}$

Ficando então:

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( x^{-\frac{1}{2}} \right)' +\left( x^{\frac{1}{2}} \right)' \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( -\frac{1}{2}\cdot x^{-\frac{1}{2}-1} \right) +\left( \frac{1}{2}\cdot x^{\frac{1}{2}-1} \right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( -\frac{1}{2}\cdot x^{-\frac{3}{2}} \right) +\left( \frac{1}{2}\cdot x^{-\frac{1}{2}} \right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( -\frac{1}{2}\cdot \frac{1}{x^{\frac{3}{2}}} \right) +\left( \frac{1}{2}\cdot \frac{1}{x^{\frac{1}{2}}} \right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \therefore f'(x)=\left( -\frac{1}{2\sqrt{x^3}} \right) +\left( \frac{1}{2\sqrt{x}} \right) \end{gathered} }$

• Substituindo no ponto f'(1), temos que:

$\large\displaystyle\text{ \begin{gathered} f'(x)=\left( -\frac{1}{2\sqrt{x^3}} \right) +\left( \frac{1}{2\sqrt{x}} \right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(1)=\left( -\frac{1}{2\sqrt{(1)^3}} \right) +\left( \frac{1}{2\sqrt{(1)}} \right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} f'(1)= \cancel{-\frac{1}{2}} +\cancel{\frac{1}{2}}\end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \therefore\boxed{\boxed{\green{f'(1)= 0}}}\ \checkmark\end{gathered} }$

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