Questão 1 - Determine as derivadas parciais de primeira ordem das seguintes funções:
\left.h(x, y, z)=\left(\cos ^{5}(x)+\operatorname{sec} \left(z y^{2}\right)\right)\right)^{\frac{2}{3}}.

Passo 1

O enunciado pede as derivadas parciais de primeira ordem de
h(x,y,z)=\bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{\tfrac{2}{3}}.
O conteúdo aqui é a regra da cadeia em várias variáveis: se definirmos
u(x,y,z)=\cos^{5}(x)+\operatorname{sec}(z\,y^{2}),
então
h(x,y,z)=u(x,y,z)^{\tfrac{2}{3}}
e usaremos
\frac{\partial h}{\partial w} =\frac{2}{3}\,u^{-1/3}\,\frac{\partial u}{\partial w},
para w=x,y,z.

Passo 2

Vamos aplicar a regra da cadeia de forma geral: para cada variável w\in\{x,y,z\},
\frac{\partial h}{\partial w} =\frac{2}{3}\,\bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}} \;\frac{\partial}{\partial w}\bigl[\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr].

Passo 3

Derivada em relação a x.
\frac{\partial}{\partial x}\bigl[\cos^{5}(x)\bigr] =5\,\cos^{4}(x)\cdot(-\sin(x)) =-5\,\cos^{4}(x)\,\sin(x),
e \operatorname{sec}(z\,y^{2}) é constante em x. Logo
h_{x}=-\frac{10}{3}\,\cos^{4}(x)\,\sin(x)\, \bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}}.

Passo 4

Derivada em relação a y.
\frac{\partial}{\partial y}\bigl[\operatorname{sec}(z\,y^{2})\bigr] =\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\cdot\frac{\partial}{\partial y}(z\,y^{2}) =2\,y\,z\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2}),
e \cos^{5}(x) é constante em y. Logo
h_{y} =\frac{4}{3}\,y\,z\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\, \bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}}.

Passo 5

Derivada em relação a z.
\frac{\partial}{\partial z}\bigl[\operatorname{sec}(z\,y^{2})\bigr] =\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\cdot\frac{\partial}{\partial z}(z\,y^{2}) =y^{2}\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2}),
e \cos^{5}(x) é constante em z. Logo
h_{z} =\frac{2}{3}\,y^{2}\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\, \bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}}.

Resposta

h_{x}=-\dfrac{10}{3}\,\cos^{4}(x)\,\sin(x)\,\bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}} h_{y}=\dfrac{4}{3}\,y\,z\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\,\bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}} h_{z}=\dfrac{2}{3}\,y^{2}\,\operatorname{sec}(z\,y^{2})\,\tan(z\,y^{2})\,\bigl(\cos^{5}(x)+\operatorname{sec}(z\,y^{2})\bigr)^{-\tfrac{1}{3}}

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