Question

Derivada parcial de primeira ordem:

f(x,y) = y² . ln(x²+y²)

Answer 1

$f(x,\,y)=y^2\,\mathrm{\ell n}(x^2+y^2)$


$\bullet\;\;$ Derivada parcial de $f$ em relação a $x:$

(considera $y$ como constante, e deriva em relação a $x$ usando as regras usuais de derivação)

$\dfrac{\partial f}{\partial x}(x,\,y)=\dfrac{\partial f}{\partial x}\big[y^2\,\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ \dfrac{\partial f}{\partial x}(x,\,y)=y^2\cdot \dfrac{\partial f}{\partial x}\big[\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ \dfrac{\partial f}{\partial x}(x,\,y)=y^2\cdot \left[\dfrac{1}{x^2+y^2}\cdot \dfrac{\partial f}{\partial x}(x^2+y^2) \right ]\\\\\\ \dfrac{\partial f}{\partial x}(x,\,y)=y^2\cdot \left[\dfrac{1}{x^2+y^2}\cdot (2x+0) \right ]\\\\\\ \therefore~~\boxed{\begin{array}{c}\dfrac{\partial f}{\partial x}(x,\,y)=\dfrac{2xy^2}{x^2+y^2} \end{array}}$


$\bullet\;\;$ Derivada parcial de $f$ em relação a $y:$

(agora, considera $x$ como constante, e deriva em relação a $y$ )

$\dfrac{\partial f}{\partial y}(x,\,y)=\dfrac{\partial f}{\partial y}\big[y^2\,\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ \dfrac{\partial f}{\partial y}(x,\,y)=\dfrac{\partial f}{\partial y}(y^2)\cdot \mathrm{\ell n}(x^2+y^2)+y^2\cdot \dfrac{\partial f}{\partial y}\big[\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ \dfrac{\partial f}{\partial y}(x,\,y)=2y\cdot \mathrm{\ell n}(x^2+y^2)+y^2\cdot \left[\dfrac{1}{x^2+y^2}\cdot \dfrac{\partial f}{\partial y}(x^2+y^2)\right]\\\\\\ \dfrac{\partial f}{\partial y}(x,\,y)=2y\,\mathrm{\ell n}(x^2+y^2)+y^2\cdot \left[\dfrac{1}{x^2+y^2}\cdot (0+2y)\right]$

$\therefore~~\boxed{\begin{array}{c}\dfrac{\partial f}{\partial y}(x,\,y)=2y\,\mathrm{\ell n}(x^2+y^2)+\dfrac{2y^3}{x^2+y^2} \end{array}}$