Question

CALCULO DE FUNÇÕES DE VÁRIAS VARIÁVEIS

Answer 1

Para derivar parcialmente, basta derivar a função considerando y constante, ou seja derivando só o x e depois derivar o y considerando x constante.

item a)

$\displaystyle \text{f(x,y)}=2\text x^{8}\text y^2-\text{x.y}^2+3\text x+1 \\\\ \underline{\text{Derivando parcialmente}}: \\\\ \frac{\partial}{\partial \text x}(2\text x^{8}\text y^2-\text{x.y}^2+3\text x+1 ) = 16\text x^7\text y^2-\text y^2+3\\\\ \frac{\partial}{\partial \text y}(2\text x^{8}\text y^2-\text{x.y}^2+3\text x+1 ) = 4\text x^8\text y-2\text{x.y} \\\\\\\ \underline{\text{Portanto}}:$

$\displaystyle \huge\boxed{\frac{\partial}{\partial \text x}[\text{f(x,y)}] = 16\text x^7\text y^2-\text y^2+3\ }\checkmark \\\\\\\ \huge\boxed{\frac{\partial}{\partial \text y} = 4\text x^8\text y -2\text{x.y}\ }\checkmark$

item b)

$\displaystyle \text{f(x,y)} = e^{\text y}+\text {y sen(xy)} \\\\ \underline{\text{Derivando parcialmente}}: \\\\ \frac{\partial}{\partial \text x} (\text e^{\text y }+\text{y sen(xy)}) = 0+\text y\text{cos(xy).y} \to \text y^2\text{cos(xy)} \\\\\\ \frac{\partial}{\partial \text y} (\text e^{\text y}+\text{y sen(xy)}) = \text e^{\text y}+[\text y]'\text{sen(xy)}+\text y[\text{sen(xy)}]'$

$\displaystyle \frac{\partial }{\partial \text y}(\text e^{\text y}+\text{y sen(xy)}) = \text e^{\text y}+\text{sen(xy)}+\text{y cos(xy) x} \\\\\\ \frac{\partial}{\partial \text y}(\text e^{\text y}+\text{y sen(xy)}) = \text e^{\text y}+\text{sen(\text xy)}+\text{xy cos(xy)}$

Portanto :

$\huge\boxed{\frac{\partial }{\partial \text x} [\ \text{f(x,y)}\ ] = \text y^2\text{cos(xy)}\ }\checkmark \\\\\\ \huge\boxed{\frac{\partial }{\partial \text x} [\ \text{f(x,y)}\ ] = \text e^{\text y}+\text{sen(xy)}+\text{xycos(xy)}}$