Question

Calcular as derivadas parciais de 2ª ordem das funções:
a) f(x,y)=x+y+xy
b) f (x, y) = x2 y + cos y + ysenx

Answer 1

$f(x,y) = x+y+x*y\\\\ \frac{df}{dx}=1+0+1*y\to \boxed{\frac{df}{dx}=1+y}\\\\ \boxed{\frac{d^2f}{dx^2} = 0+0}\\\\ \\\ f(x,y) = x+y+x*y\\\\ \boxed{\frac{df}{dy}= 0+1+x*1 = 1+x}\\\\\\ \boxed{\frac{d^2f}{dy^2} =0+0}$

B)

$f(x,y)= 2x^2*y + cos(y)+y*sen(x)$

derivando pra x 
$\frac{\partial f}{\partial x} =2x*y + 0 + y*cos(x) \\\\ \boxed{ \frac{\partial f}{\partial x} =2xy +y*cos(x)}$

derivando novamente pra x
$\frac{\partial ^2 f}{\partial x^2} =2*1*y + y*-sen(x)\\\\ \boxed{ \frac{\partial ^2 f}{\partial x^2} =2y-y*sen(x)}$

derivando em relação a y
$f(x,y)= 2x^2*y + cos(y)+y*sen(x)$

$\frac{\partial f}{\partial y}= 2x^2*1-sen(y)+1*sen(x)\\\\ \boxed{\frac{\partial f}{\partial y}=2x^2-sen(y)+sen(x)}$

derivando novamente em relação a y
$\frac{\partial^2 f}{\partial y^2}= 0-cos(y)+0\\\\ \boxed{\frac{\partial^2 f}{\partial y^2}=-cos(y)}$