Determine as derivadas parciais de primeira ordem das funções definidas por
L(x, y, z, w)=\operatorname{sec} \left(-2 x y z w^{2}\right) \frac{\ln \left(w^{2}-2 z y\right)}{\cos \left(x+y-z+w^{2}\right)}.

Passo 1

O enunciado quer as derivadas parciais de primeira ordem de
L(x,y,z,w)=\operatorname{sec}\bigl(-2\,x\,y\,z\,w^{2}\bigr)\,\frac{\ln\bigl(w^{2}-2\,z\,y\bigr)}{\cos\bigl(x+y-z+w^{2}\bigr)}.
O conteúdo principal é derivada parcial em função de várias variáveis, aplicando regra do produto e regra da cadeia (além de derivada de \operatorname{sec}, \ln e \cos).

Passo 2

Definimos
u=-2\,x\,y\,z\,w^{2},\quad A=\operatorname{sec}(u),
v=x+y-z+w^{2},\quad B=\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v),
de modo que L=A\cdot B. Pela regra do produto,
\frac{\partial L}{\partial x} =A_x\,B + A\,B_x.
Calculamos
A_x=\operatorname{sec}(u)\,\tan(u)\,\frac{\partial u}{\partial x} =\operatorname{sec}(u)\,\tan(u)\,(-2\,y\,z\,w^{2}),
B_x =\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v)\,\tan(v)\,\frac{\partial v}{\partial x} =\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v)\,\tan(v).
Substituindo em \partial L/\partial x,
\frac{\partial L}{\partial x} =\operatorname{sec}(u)\,\tan(u)\,(-2yzw^{2})\;\frac{\ln(w^{2}-2zy)}{\cos(v)} \;+\;\operatorname{sec}(u)\,\ln(w^{2}-2zy)\,\operatorname{sec}(v)\,\tan(v).

Passo 3

Para \partial L/\partial y usamos o mesmo A e B.
A_y=\operatorname{sec}(u)\,\tan(u)\,\frac{\partial u}{\partial y} =\operatorname{sec}(u)\,\tan(u)\,(-2\,x\,z\,w^{2}),
B_y =\frac{-2\,z}{\,w^{2}-2\,z\,y\,}\,\operatorname{sec}(v) \;+\;\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v)\,\tan(v).
Então
\frac{\partial L}{\partial y} =A_y\,B + A\,B_y.

Passo 4

Para \partial L/\partial z:
A_z=\operatorname{sec}(u)\,\tan(u)\,(-2\,x\,y\,w^{2}),
B_z =\frac{-2\,y}{\,w^{2}-2\,z\,y\,}\,\operatorname{sec}(v) \;-\;\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v)\,\tan(v),
\frac{\partial L}{\partial z} =A_z\,B + A\,B_z.

Passo 5

Para \partial L/\partial w:
A_w=\operatorname{sec}(u)\,\tan(u)\,\frac{\partial u}{\partial w} =\operatorname{sec}(u)\,\tan(u)\,(-4\,x\,y\,z\,w),
B_w =\frac{2\,w}{\,w^{2}-2\,z\,y\,}\,\operatorname{sec}(v) \;+\;2\,w\,\ln\bigl(w^{2}-2\,z\,y\bigr)\,\operatorname{sec}(v)\,\tan(v),
\frac{\partial L}{\partial w} =A_w\,B + A\,B_w.

Resposta

\frac{\partial L}{\partial x} =\operatorname{sec}(u)\,\tan(u)\,(-2yzw^{2})\;\frac{\ln(w^{2}-2zy)}{\cos(v)} \;+\;\operatorname{sec}(u)\,\ln(w^{2}-2zy)\,\operatorname{sec}(v)\,\tan(v),
\frac{\partial L}{\partial y} =\operatorname{sec}(u)\,\tan(u)\,(-2xzw^{2})\;\frac{\ln(w^{2}-2zy)}{\cos(v)} \;+\;\operatorname{sec}(u)\Bigl[\frac{-2z}{\,w^{2}-2zy\,}\operatorname{sec}(v) +\ln(w^{2}-2zy)\operatorname{sec}(v)\tan(v)\Bigr],
\frac{\partial L}{\partial z} =\operatorname{sec}(u)\,\tan(u)\,(-2xyw^{2})\;\frac{\ln(w^{2}-2zy)}{\cos(v)} \;+\;\operatorname{sec}(u)\Bigl[\frac{-2y}{\,w^{2}-2zy\,}\operatorname{sec}(v) -\ln(w^{2}-2zy)\operatorname{sec}(v)\tan(v)\Bigr],
\frac{\partial L}{\partial w} =\operatorname{sec}(u)\,\tan(u)\,(-4xyz\,w)\;\frac{\ln(w^{2}-2zy)}{\cos(v)} \;+\;\operatorname{sec}(u)\Bigl[\frac{2w}{\,w^{2}-2zy\,}\operatorname{sec}(v) +2w\,\ln(w^{2}-2zy)\operatorname{sec}(v)\tan(v)\Bigr],
onde u=-2xyz\,w^{2} e v=x+y-z+w^{2}.

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