Question

qual a derivada parcial da f(x,y)=y^2*ln(x^2 +y^2)

Answer 1

$\dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial x}\left[y^2\ln(x^2+y^2)\right]\\ \\ \\ \dfrac{\partial f}{\partial x}=y^2\cdot\dfrac{\partial }{\partial x}\left[\ln(x^2+y^2)\right]\\ \\ \\ \dfrac{\partial f}{\partial x}=y^2\cdot\dfrac{\dfrac{\partial }{\partial x}(x^2+y^2)}{x^2+y^2}$

$\dfrac{\partial f}{\partial x}=y^2\cdot\dfrac{\dfrac{\partial }{\partial x}(x^2)+\dfrac{\partial }{\partial x}(y^2)}{x^2+y^2}\\ \\ \\ \dfrac{\partial f}{\partial x}=y^2\cdot\dfrac{2x+0}{x^2+y^2}\\ \\ \\ \boxed{\dfrac{\partial f}{\partial x}=\dfrac{2xy^2}{x^2+y^2}}$

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$\dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}\left[y^2\ln(x^2+y^2)\right]\\ \\ \\ \dfrac{\partial f}{\partial y}=\left[\dfrac{\partial }{\partial y}(y^2)\right]\cdot\ln(x^2+y^2)+y^2\cdot \dfrac{\partial }{\partial y}\ln(x^2+y^2)\\ \\ \\ \dfrac{\partial f}{\partial y}=2y\cdot\ln(x^2+y^2)+y^2\cdot \dfrac{\dfrac{\partial }{\partial y}(x^2+y^2)}{x^2+y^2}$

$\dfrac{\partial f}{\partial y}=2y\cdot\ln(x^2+y^2)+y^2\cdot \dfrac{2y}{x^2+y^2}\\ \\ \\ \boxed{\dfrac{\partial f}{\partial y}=2y\cdot\ln(x^2+y^2)+\dfrac{2y^3}{x^2+y^2}}$