Question

Qual a derivada parcial de f(x,y)= √ (x^2+y^2)

Ajuda ai, Obrigado!

Answer 1

$f\left(x,\,y \right )=\sqrt{x^{2}+y^{2}}$


$\bullet\;\;\dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left(\sqrt{x^{2}+y^{2}} \right )\\ \\ \dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left[\left(x^{2}+y^{2} \right )^{\,^{1}\!\!\!\diagup\!\!_{2}} \right ]\\ \\ \dfrac{\partial f}{\partial x}=\dfrac{1}{2}\cdot \left(x^{2}+y^{2} \right )^{\,^{1}\!\!\!\diagup\!\!_{2}-1}\cdot \dfrac{\partial}{\partial x}\left(x^{2}+y^{2} \right )\\ \\ \dfrac{\partial f}{\partial x}=\dfrac{1}{2}\cdot \left(x^{2}+y^{2} \right )^{\,^{-1}\!\!\!\diagup\!\!_{2}}\cdot\left(2x+0 \right )\\ \\ \dfrac{\partial f}{\partial x}=\dfrac{1}{\diagup\!\!\!\! 2\sqrt{x^{2}+y^{2}}}\cdot \diagup\!\!\!\! 2x\\ \\ \boxed{\dfrac{\partial f}{\partial x}=\dfrac{x}{\sqrt{x^{2}+y^{2}}}}$


$\bullet\;\;\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left(\sqrt{x^{2}+y^{2}} \right )\\ \\ \dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left[\left(x^{2}+y^{2} \right )^{\,^{1}\!\!\!\diagup\!\!_{2}} \right ]\\ \\ \dfrac{\partial f}{\partial y}=\dfrac{1}{2}\cdot \left(x^{2}+y^{2} \right )^{\,^{1}\!\!\!\diagup\!\!_{2}-1}\cdot \dfrac{\partial}{\partial y}\left(x^{2}+y^{2} \right )\\ \\ \dfrac{\partial f}{\partial y}=\dfrac{1}{2}\cdot \left(x^{2}+y^{2} \right )^{\,^{-1}\!\!\!\diagup\!\!_{2}}\cdot\left(0+2y \right )\\ \\ \dfrac{\partial f}{\partial y}=\dfrac{1}{\diagup\!\!\!\! 2\sqrt{x^{2}+y^{2}}}\cdot \diagup\!\!\!\! 2y\\ \\ \boxed{\dfrac{\partial f}{\partial y}=\dfrac{y}{\sqrt{x^{2}+y^{2}}}}$