Question

Como resolve esses exercícios de cálculo?

Answer 1

a) $f\left(x,\,y \right )=e^{x^{2}+y^{2}-4}$

$\bullet\;\;\dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left(e^{x^{2}+y^{2}-4} \right )\\ \\ \dfrac{\partial f}{\partial x}=e^{x^{2}+y^{2}-4}\cdot\dfrac{\partial}{\partial x}\left(x^{2}+y^{2}-4 \right )\\ \\ \boxed{\dfrac{\partial f}{\partial x}=e^{x^{2}+y^{2}-4}\cdot 2x}\\ \\ \\ \bullet\;\;\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left(e^{x^{2}+y^{2}-4} \right )\\ \\ \dfrac{\partial f}{\partial y}=e^{x^{2}+y^{2}-4}\cdot\dfrac{\partial}{\partial y}\left(x^{2}+y^{2}-4 \right )\\ \\ \boxed{\dfrac{\partial f}{\partial y}=e^{x^{2}+y^{2}-4}\cdot 2y}$


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

$\bullet\;\;\dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left[\,e^{x^{2}}\left(x^{2}+y^{2} \right )\, \right ]\\ \\ \dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left(e^{x^{2}} \right )\cdot \left(x^{2}+y^{2} \right )+e^{x^{2}}\cdot\dfrac{\partial}{\partial x}\left(x^{2}+y^{2} \right )\\ \\ \dfrac{\partial f}{\partial x}=e^{x^{2}}\cdot\dfrac{\partial}{\partial x}\left(x^{2} \right )\cdot \left(x^{2}+y^{2} \right )+e^{x^{2}}\cdot 2x\\ \\ \boxed{\dfrac{\partial f}{\partial x}=e^{x^{2}}\cdot 2x\cdot \left(x^{2}+y^{2} \right )+e^{x^{2}}\cdot 2x}$


$\bullet\;\;\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left[\,e^{x^{2}}\left(x^{2}+y^{2} \right )\, \right ]\\ \\ \dfrac{\partial f}{\partial y}=e^{x^{2}}\cdot \dfrac{\partial}{\partial y}\left(x^{2}+y^{2} \right )\\ \\ \boxed{\dfrac{\partial f}{\partial y}=e^{x^{2}}\cdot 2y}$


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

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


$\bullet\;\;\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left[\,y^{2}\mathrm{\,\ell n}\left(x^{2}+y^{2}\right)\, \right ]\\ \\ \dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left(y^{2} \right )\cdot \mathrm{\ell n}\left(x^{2}+y^{2} \right )+y^{2}\cdot \dfrac{\partial}{\partial y}\left[\,\mathrm{\ell n}\left(x^{2}+y^{2}\right)\, \right ]\\ \\ \dfrac{\partial f}{\partial y}=2y\cdot \mathrm{\ell n}\left(x^{2}+y^{2} \right )+y^{2}\cdot \dfrac{1}{x^{2}+y^{2}}\cdot \dfrac{\partial}{\partial y}\left(x^{2}+y^{2}\right)\\ \\ \dfrac{\partial f}{\partial y}=2y\mathrm{\,\ell n}\left(x^{2}+y^{2} \right )+y^{2}\cdot \dfrac{1}{x^{2}+y^{2}}\cdot 2y\\ \\ \boxed{\dfrac{\partial f}{\partial y}=2y\mathrm{\,\ell n}\left(x^{2}+y^{2} \right )+\dfrac{2y^{3}}{x^{2}+y^{2}}}$


d) $f\left(x,\,y,\,z \right )=x^{2}yz-xy$

$\bullet\;\;\dfrac{\partial f}{\partial x}=\dfrac{\partial}{\partial x}\left(x^{2}yz-xy \right )\\ \\ \dfrac{\partial f}{\partial x}=yz\cdot \dfrac{\partial}{\partial x}\left(x^{2} \right )-y\cdot \dfrac{\partial}{\partial x}\left(x \right )\\ \\ \dfrac{\partial f}{\partial x}=yz\cdot 2x-y\cdot 1\\ \\ \boxed{\dfrac{\partial f}{\partial x}=2xyz-y}$


$\bullet\;\;\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}\left(x^{2}yz-xy \right )\\ \\ \dfrac{\partial f}{\partial y}=x^{2}z\cdot \dfrac{\partial}{\partial y}\left(y \right )-x\cdot \dfrac{\partial}{\partial y}\left(y \right )\\ \\ \dfrac{\partial f}{\partial y}=x^{2}z\cdot 1-x\cdot 1\\ \\ \boxed{\dfrac{\partial f}{\partial y}=x^{2}z-x}$


$\bullet\;\;\dfrac{\partial f}{\partial z}=\dfrac{\partial}{\partial z}\left(x^{2}yz-xy \right )\\ \\ \dfrac{\partial f}{\partial z}=x^{2}y\cdot \dfrac{\partial}{\partial z}\left(z \right )- \dfrac{\partial}{\partial z}\left(xy \right )\\ \\ \dfrac{\partial f}{\partial z}=x^{2}y\cdot 1- 0\\ \\ \boxed{\dfrac{\partial f}{\partial z}=x^{2}y}$