Achar as derivadas parciais das funções:
z=\ln \operatorname{sen} \frac{x+a}{\sqrt{y}}.

Passo 1

O enunciado pede as derivadas parciais de
z = \ln\bigl(\operatorname{sen}\!\bigl(\tfrac{x+a}{\sqrt{y}}\bigr)\bigr).
O conteúdo principal é a regra da cadeia aplicada duas vezes: primeiro em \ln(u) e depois em \operatorname{sen}(v), além da derivada de v=(x+a)y^{-1/2}.

Passo 2

Calcular \displaystyle \frac{\partial z}{\partial x}. Seja
u = \operatorname{sen}(v), \quad v = \frac{x+a}{\sqrt{y}}.
Então usando \frac{d}{dx}\ln(u)=\frac1u\,\frac{du}{dx} e \frac{d}{dx}\operatorname{sen}(v)=\cos(v)\,\frac{dv}{dx}, temos
\frac{\partial z}{\partial x} = \frac{1}{u}\,\frac{\partial u}{\partial x} = \frac{1}{\operatorname{sen}(v)}\;\cos(v)\;\frac{\partial }{\partial x}\!\Bigl(\frac{x+a}{\sqrt{y}}\Bigr).
Como \frac{\partial }{\partial x}\bigl((x+a)/\sqrt{y}\bigr)=1/\sqrt{y}, resulta
\frac{\partial z}{\partial x} = \frac{\cos(v)}{\operatorname{sen}(v)}\;\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{y}}\;\cot(v).

Passo 3

Calcular \displaystyle \frac{\partial z}{\partial y}. De novo \displaystyle \frac{\partial z}{\partial y}=\frac{1}{u}\,\frac{\partial u}{\partial y} com \displaystyle \frac{\partial u}{\partial y}=\cos(v)\,\frac{\partial v}{\partial y}. Agora
\frac{\partial }{\partial y}\!\Bigl(\frac{x+a}{\sqrt{y}}\Bigr) = (x+a)\;\frac{\partial }{\partial y}\bigl(y^{-1/2}\bigr) = (x+a)\,\Bigl(-\tfrac12\,y^{-3/2}\Bigr) = -\frac{x+a}{2\,y^{3/2}}.
Logo
\frac{\partial z}{\partial y} = \frac{1}{\operatorname{sen}(v)}\;\cos(v)\;\Bigl(-\frac{x+a}{2\,y^{3/2}}\Bigr) = -\frac{x+a}{2\,y^{3/2}}\;\cot(v).

Resposta

\frac{\partial z}{\partial x} = \frac{1}{\sqrt{y}}\;\cot\!\Bigl(\frac{x+a}{\sqrt{y}}\Bigr), \qquad \frac{\partial z}{\partial y} = -\,\frac{x+a}{2\,y^{3/2}}\;\cot\!\Bigl(\frac{x+a}{\sqrt{y}}\Bigr).

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