Question

O valor das derivadas parciais de 1ª e 2ª ordem da função dada por z = cos 2xy.
a)-2x Sen 2xy; -4x² cos 2xy; -4xy cos 2xy -2 sen 2xy
b)-2xy Sen 2xy; -4xy² cos 2xy; -4xy cos 2xy+2 sen 2xy
c)2xy Sen 2xy; -4xy² cos 2xy; -4xy cos 2xy + 2 sen 2xy
d)2y Sen 2xy; 4y² cos 2xy; + 4xy cos 2 xy+ 2 Sen 2xy
e)-2y Sen 2xy; -4y² cos 2xy; -4xy cos 2xy -2 sen 2xy

Answer 1

Resposta:

$\sf z=cos\,2xy$

Derivada parcial de 1ª ordem em relação a x:

(Trate a variável y como uma constante.)

$\sf \dfrac{\partial z}{\partial x}=\dfrac{\partial}{\partial x}(cos\,2xy)$

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

$\sf \dfrac{\partial z}{\partial x}=\dfrac{\partial}{\partial u}(cos\,u)\cdot \dfrac{\partial}{\partial x}(2xy)$

$\sf \dfrac{\partial z}{\partial x}=(-\,2sen\,u)\cdot \dfrac{\partial}{\partial x}(2x)\cdot y$

$\sf \dfrac{\partial z}{\partial x}=(-\,2sen\,u)\cdot 2y$

$\sf \dfrac{\partial z}{\partial x}=-\,2y(sen\,u)$

Substituindo de volta u = 2xy:

$\red{\boxed{\sf \dfrac{\partial z}{\partial x}=-\,2y(sen\,2xy)}}$

Derivada pura:

$\sf \dfrac{\partial^2 z}{\partial x^2}=\dfrac{\partial}{\partial x}\bigg(\dfrac{\partial z}{\partial x}\bigg)$

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

$\sf \dfrac{\partial^2 z}{\partial x^2}=-\,2y\dfrac{\partial}{\partial x}(sen\,2xy)$

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

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

$\sf \dfrac{\partial^2 z}{\partial x^2}=-\,2y(cos\,u)\cdot2y$

$\sf \dfrac{\partial^2 z}{\partial x^2}=-\,4y^2(cos\,u)$

Substituindo de volta u = 2xy:

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

Derivada mista:

$\sf \dfrac{\partial^2 z}{\partial y\,\partial x}=\dfrac{\partial}{\partial y}\bigg(\dfrac{\partial z}{\partial x}\bigg)$

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

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

Aplicando a regra do produto:

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

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

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

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:

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

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

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

Substituindo de volta u = 2xy:

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

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

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

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

Letra E

OBS.:

Regra da cadeia:

$\sf f'(g)=f'(u)\cdot g'$

Regra do produto:

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