Matemática, perguntado por aryeleazevedoary, 1 ano atrás

Derivada implícita : raiz (x+y) + raiz(x-y)=6

Soluções para a tarefa

Respondido por ArthurPDC
0
Derivando os dois lados da expressão em relação a x:

\sqrt{x+y}+\sqrt{x-y}=6\\\\
\dfrac{d}{dx}(\sqrt{x+y}+\sqrt{x-y})=\dfrac{d}{dx}(6)\\\\
\dfrac{d}{dx}(\sqrt{x+y})+\dfrac{d}{dx}(\sqrt{x-y})=0\\\\
\dfrac{d}{dx}[(x+y)^{\frac 1 2}]+\dfrac{d}{dx}[(x-y)^{\frac 1 2}]=0\\\\
~[\frac 1 2 \cdot (x+y)^{-\frac 1 2}\cdot(1+y')]+[\frac 1 2 \cdot (x-y)^{-\frac 1 2}\cdot(1-y')]=0\\\\
~[(x+y)^{-\frac 1 2}\cdot(1+y')]+[(x-y)^{-\frac 1 2}\cdot(1-y')]=0\\\\
(x+y)^{-\frac 1 2}+y'(x+y)^{-\frac 1 2}+(x-y)^{-\frac 1 2}-y'(x-y)^{-\frac 1 2}=0

y'[(x-y)^{-\frac 1 2}-(x+y)^{-\frac 1 2}]=(x+y)^{-\frac 1 2}+(x-y)^{-\frac 1 2}\\\\
y'=\dfrac{(x+y)^{-\frac 1 2}+(x-y)^{-\frac 1 2}}{(x-y)^{-\frac 1 2}-(x+y)^{-\frac 1 2}}\\\\
y'=\dfrac{\dfrac{1}{\sqrt{x+y}}+\dfrac{1}{\sqrt{x-y}}}{\dfrac{1}{\sqrt{x-y}}-\dfrac{1}{\sqrt{x+y}}}\\\\\\
y'=\dfrac{~~~\dfrac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x-y}\cdot\sqrt{x+y}}~~~}{\dfrac{\sqrt{x+y}-\sqrt{x-y}}{\sqrt{x-y}\cdot\sqrt{x+y}}}\\\\\\
y'=\dfrac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}

Podemos racionalizar a resposta encontrada acima:

y'=\dfrac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}\times\dfrac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}\\\\
y'=\dfrac{(\sqrt{x+y}+\sqrt{x-y})^2}{(\sqrt{x+y})^2-(\sqrt{x-y})^2}\\\\
y'=\dfrac{(\sqrt{x+y})^2+2\cdot\sqrt{x+y}\cdot\sqrt{x-y}+(\sqrt{x-y})^2}{(x+y)-(x-y)}\\\\
y'=\dfrac{(x+y)+2\sqrt{x^2-y^2}+(x-y)}{(x+y)-(x-y)}\\\\
y'=\dfrac{2x+2\sqrt{x^2-y^2}}{2y}\\\\
\boxed{y'=\dfrac{x+\sqrt{x^2-y^2}}{y}}
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