Question

Encontre as derivadas parciais de segunda ordem dá função:
F(x, y) = 4x^3y^4 + 4x - 6y^5

Alguém pode ajudar por favor??

Answer 1

Olá

$\mathsf{f(x,y)=4x^3y^4+4x-6y^5}$

Derivando em relação a x.

$\displaystyle \mathsf{ \frac{\partial f(x,y)}{\partial x} =\frac{\partial }{\partial x}(4x^3y^4)+\frac{\partial }{\partial x}(4x)-\frac{\partial }{\partial x}(6y^5)}\\\\\\\mathsf{ \frac{\partial f(x,y)}{\partial x} =12x^2y^4~+~4~-0}\\\\\\\boxed{\mathsf{ \frac{\partial f(x,y)}{\partial x} =12x^2y^4~+~4~}}$

Derivada de segunda ordem em relação a x

$\displaystyle \mathsf{ \frac{\partial ^2f(x,y)}{\partial x^2} =\frac{\partial }{\partial x}(12x^2y^4)+\frac{\partial }{\partial x}(4)}\\\\\\\mathsf{ \frac{\partial^2 f(x,y)}{\partial x^2} =24xy^4+0}\\\\\\\boxed{\mathsf{ \frac{\partial ^2f(x,y)}{\partial x^2} =24xy^4}}$




Derivando em relação a y

$\displaystyle \mathsf{ \frac{\partial f(x,y)}{\partial y} =\frac{\partial }{\partial y}(4x^3y^4)+\frac{\partial }{\partial y}(4x)-\frac{\partial }{\partial y}(6y^5)}\\\\\\\mathsf{ \frac{\partial f(x,y)}{\partial y} =16x^3y^3~+~0~-30y^4}\\\\\\\boxed{\mathsf{ \frac{\partial f(x,y)}{\partial y} =16x^3y^3~-~30^4~}}$

Derivada de segunda ordem em relação a y

$\displaystyle \mathsf{ \frac{\partial^2 f(x,y)}{\partial y^2} =\frac{\partial }{\partial y}(16x^3y^3)-\frac{\partial }{\partial y}(30y^4)}\\\\\\\boxed{\mathsf{ \frac{\partial^2 f(x,y)}{\partial y^2} =48x^3y^2-120y^3}}$