Question

Encontre a derivada das seguintes funções:

Answer 1

Explicação passo-a-passo:

Cálculo diferencial

Dada a função:

$\sf{ f(x)~=~\dfrac{1}{\sqrt{2x-1}} }$

Vamos reescrever a função:

$\iff \sf{ y~=~ \left(\dfrac{1}{2x-1}\right)^{\frac{1}{2}} }$

Vamos aplicar logaritmos em ambos os membros da nossa igualdade e nos apoderar d'algumas propriedades dos mesmos .

$\iff\sf{ \ln(y)~=~\dfrac{1}{2}\ln\left[\dfrac{1}{2x-1}\right] }$

$\iff\sf{ \ln(y)~=~\dfrac{1}{2}\ln(1)-\dfrac{1}{2}\ln\left(2x - 1\right) }$

Olha que $\sf{\left[\ln(u)\right]~=~\dfrac{u'}{u}}$ , vamos derivar em ambos os membros da nossa igualdade :

$\iff\sf{ \dfrac{y'}{y}~=~0-\dfrac{1}{2}\dfrac{(2x-1)'}{2x-1}~=~-\dfrac{1}{2x-1} }$

$\iff\sf{ y'~=~-y*\dfrac{1}{2x-1} }$

$\iff\sf{ y'~=~-\dfrac{1}{\left(2x-1\right)\sqrt{2x-1}} }$

$\iff \boxed{\boxed{ \green{\sf{ y'~=~-\dfrac{1}{\left(2x-1\right)\sqrt{2x-1}} } } } }$

=================================================================

Vamos cá também fazer a mesma ginástica.

$\sf{f(x)~=~\sqrt[3]{x+3} }$

$\iff \sf{ y~=~\left( x+3\right)^{\frac{1}{3}} }$

$\iff \sf{ \ln(y)~=~\dfrac{1}{3}\ln(x+3) }$

$\iff \sf{ \dfrac{y'}{y}~=~ \dfrac{1}{3}*\dfrac{(x+3)'}{x+3} }$

$\iff \sf{ y'~=~y\dfrac{1}{3(x+3)} }$

$\iff\boxed{\boxed{\green{\sf{ f'(x)~=~\dfrac{\sqrt[3]{x+3}}{3(x+3)} } } } }$

This answer was elaborad by:

Murrima , Joaquim Marcelo

UEM(MOÇAMBIQUE)-DMI

Answer 2

• Calculando as derivadas, temos:

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

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

Desejamos calcular a derivada das seguintes funções:

$\large\displaystyle\text{ \begin{gathered} f(x)=\frac{1}{\sqrt{2x-1} } \end{gathered} }$

$\large\displaystyle\begin{gathered} f(x)=\sqrt[3]{x+3} \end{gathered}$

Para calcular derivadas, temos algumas propriedades de derivação. Sendo elas:

$\large\displaystyle\begin{gathered} \left( f\cdot g \right)' = f'\cdot g+f\cdot g'\ \ \ \ \ \bf (I)\end{gathered}$

$\large\displaystyle\begin{gathered} \left( \frac{f}{g} \right)' = \frac{f'\cdot g-f\cdot g'}{g^2}\ \ \ \ \ \bf (II)\end{gathered}$

$\large\displaystyle\begin{gathered} f\left( g \left( x \right) \right) '= f'(g(x))\cdot g'(x)\ \ \ \ \ \bf (III)\end{gathered}$

$\large\displaystyle\begin{gathered} f(x)=a\cdot x^n\rightarrow a\cdot n\cdot x^{n-1}\ \ \ \ \ \bf (IV)\end{gathered}$

Calculando então a derivada da primeira função:

     $\large\displaystyle\text{ \begin{gathered} f'(x)=\left(2x-1 \right)^{ -\frac{1}{2}} \Rightarrow -\frac{1}{2}\left( 2x-1\right)^{-\frac{1}{2} - 1}\cdot (2x-1)' \end{gathered} }$ $\large\displaystyle\text{ \begin{gathered} f'(x)= -\frac{1}{2}\left( 2x-1\right)^{-\frac{1}{2} - 1}\cdot (2x-1)'\Rightarrow f'(x)=-\frac{(2x-1)^{-\frac{3}{2}}}{\not{2}}\cdot \not{2 }\end{gathered} }$

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

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

  $\large\displaystyle\text{ \begin{gathered} \therefore f(x)=\frac{1}{\sqrt{2x-1} }\Rightarrow \boxed{\boxed{\green{f'(x)=-\frac{1}{\sqrt{(2x-1)^3}}}}}\ \checkmark \end{gathered} }$

Calculando agora a derivada da segunda função:

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

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

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

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