Question

Dada $x^{1/2} +y^{1/2}=2$ , mostre que $\frac{d^{2}y }{dx^{2} } =\frac{1}{x^{3/2} }$

Answer 1

Olá! tudo bom?

Para monstrar que a derivada de segunda ordem de y é igual a 1/√x³, devemos primeiramente isolar y na eq dada. E para isso, temos que:

$\large\displaystyle\text{ \begin{gathered} x^{\frac{1}{2}}+y^{\frac{1}{2} }=2\end{gathered} }$

Que é a mesma coisa que:

$\large\displaystyle\begin{gathered} \sqrt{x}+\sqrt{y} =2\end{gathered}$

Isolando $\large\displaystyle\begin{gathered}\sqrt{y} \end{gathered}$ ficamos da seguinte forma:

$\large\displaystyle\begin{gathered} \sqrt{y} =2-\sqrt{x} \end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \boxed{y=\left(2-\sqrt{x}\right)^2} \end{gathered} }$

E desse modo, surge que:

$\large\displaystyle\begin{gathered} \frac{d^2y}{dx^2}=\frac{d^2}{dx^2} \left(2-\sqrt{x} \right)^2=\frac{d^2}{dx^2}\left(4-4\sqrt{x}+x\right) \end{gathered}$

Que pela lineariedade:

$\large\displaystyle\begin{gathered} \frac{d^2y}{dx^2}= \frac{d^2}{dx^2}(4)-\frac{d^2}{dx^2}\left(4\sqrt{x}\right)+\frac{d^2}{dx^2}(x)\end{gathered}$

Lembrando que a derivada de uma constante é igual a zero, logo:

$\large\displaystyle\begin{gathered} \frac{d^2y}{dx^2}= -\frac{d^2}{dx^2}\left(4\sqrt{x}\right)\end{gathered}$

$\large\displaystyle\begin{gathered} \frac{d^2y}{dx^2}= -\frac{d}{dx}\left(\frac{d}{dx}\left(4\sqrt{x}\right)\right)\end{gathered}$

Para resolver isso, basta aplicar a regra do tombo $\large\displaystyle\begin{gathered} (x^n)'=nx^{n-1}\end{gathered}$.

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= -\frac{d}{dx}\left(\frac{d}{dx}\left(4\cdot x^{\frac{1}{2} }\right)\right)\end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= -\frac{d}{dx}\left(\frac{d}{dx}\left(4\cdot \frac{1}{2}\cdot x^{\frac{1}{2}-1 }\right)\right)\end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= -\frac{d}{dx}\left(\frac{d}{dx}\left(2\cdot x^{-\frac{1}{2} }\right)\right)\end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= -\frac{d}{dx}\left(\frac{2}{\sqrt{x}} \right)\end{gathered} }$

Agora só nos resta por fim resolver essa última derivada, pra isso teremos que utilizar a derivada do quociente dada da seguinte forma:

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

Com isso, temos que:

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= -\left(\frac{2'\cdot \sqrt{x} -2\cdot (\sqrt{x} )'}{\left(\sqrt{x}\right)^2 }\right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= 2\cdot\left(\frac{ \frac{1}{2} x^{\frac{1}{2}-1 }}{x }\right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}= 2\cdot\left(\frac{ \frac{1}{2} x^{-\frac{1}{2} }}{x }\right) \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{d^2y}{dx^2}=\not2\cdot \frac{1}{\not2x^{\frac{3}{2}}} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\boxed{ \frac{d^2y}{dx^2}= \frac{1}{x^{\frac{3}{2}}}}\ \ c.q.d\ \checkmark \end{gathered} }$

Qualquer dúvida quanto a resolução dada é só chamar!

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