Resposta:

[tex]\sf z=cos\,2xy[/tex]

Derivada parcial de 1ª ordem em relação a x:

(Trate a variável y como uma constante.)

[tex]\sf \dfrac{\partial z}{\partial x}=\dfrac{\partial}{\partial x}(cos\,2xy)[/tex]

Fazendo 2xy = u e aplicando a regra da cadeia:

[tex]\sf \dfrac{\partial z}{\partial x}=\dfrac{\partial}{\partial u}(cos\,u)\cdot \dfrac{\partial}{\partial x}(2xy)[/tex]

[tex]\sf \dfrac{\partial z}{\partial x}=(-\,2sen\,u)\cdot \dfrac{\partial}{\partial x}(2x)\cdot y[/tex]

[tex]\sf \dfrac{\partial z}{\partial x}=(-\,2sen\,u)\cdot 2y[/tex]

[tex]\sf \dfrac{\partial z}{\partial x}=-\,2y(sen\,u)[/tex]

Substituindo de volta u = 2xy:

[tex]\red{\boxed{\sf \dfrac{\partial z}{\partial x}=-\,2y(sen\,2xy)}}[/tex]

Derivada pura:

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=\dfrac{\partial}{\partial x}\bigg(\dfrac{\partial z}{\partial x}\bigg)[/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=\dfrac{\partial}{\partial x}[-2y(sen\,2xy)][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=-\,2y\dfrac{\partial}{\partial x}(sen\,2xy)[/tex]

Fazendo 2xy = u e aplicando a regra da cadeia:

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=-\,2y\dfrac{\partial}{\partial u}(sen\,u)\cdot\dfrac{\partial}{\partial x}(2xy)[/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=-\,2y(cos\,u)\cdot2y[/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial x^2}=-\,4y^2(cos\,u)[/tex]

Substituindo de volta u = 2xy:

[tex]\red{\boxed{\sf \dfrac{\partial^2 z}{\partial x^2}=-\,4y^2(cos\,2xy)}}[/tex]

Derivada mista:

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=\dfrac{\partial}{\partial y}\bigg(\dfrac{\partial z}{\partial x}\bigg)[/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=\dfrac{\partial}{\partial y}[-\,2y(sen\,2xy)][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\dfrac{\partial}{\partial y}[y(sen\,2xy)][/tex]

Aplicando a regra do produto:

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\bigg[\dfrac{\partial}{\partial y}(y)\cdot (sen\,2xy)+\dfrac{\partial}{\partial y}(sen\,2xy)\cdot y\bigg][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\bigg[1\cdot (sen\,2xy)+\dfrac{\partial}{\partial y}(sen\,2xy)\cdot y\bigg][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\bigg[sen\,2xy+\dfrac{\partial}{\partial y}(sen\,2xy)\cdot y\bigg][/tex]

Pra calcular a derivada de sen 2xy em relação a y segue o mesmo raciocínio das outras vezes que aplicamos a regra da cadeia, a diferença é que agora x deve ser tratado como uma constante. Fazendo u = 2xy:

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\bigg[sen\,2xy+\dfrac{\partial}{\partial y}(sen\,u)\cdot\dfrac{\partial}{\partial y}(2xy)\cdot y\bigg][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\bigg[sen\,2xy+(cos\,u)\cdot\dfrac{\partial}{\partial y}(2y)\cdot x\cdot y\bigg][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\big[sen\,2xy+(cos\,u)\cdot2xy\big][/tex]

Substituindo de volta u = 2xy:

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2\big[sen\,2xy+(cos\,2xy)\cdot2xy\big][/tex]

[tex]\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,2sen\,2xy-4xy(cos\,2xy)[/tex]

[tex]\red{\boxed{\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=-\,4xy(cos\,2xy)-\,2(sen\,2xy)}}[/tex]

Sendo assim a ordem é: - 2y(sen 2xy); - 4y²(cos 2xy); - 4xy cos(2xy) - 2 sen(2xy)

Letra E

OBS.:

Regra da cadeia:

[tex]\sf f'(g)=f'(u)\cdot g'[/tex]

Regra do produto:

[tex]\sf [f(x)\cdot g(x)]'=f'(x)\cdot g(x)+g'(x)\cdot f(x)[/tex]