Questão 11. Calcular as derivadas parciais das seguintes funções:
f(x, y)=\frac{\operatorname{sen}\left(\mathrm{e}^{x^{2} y}\right)}{\ln (x y)}

Passo 1

O enunciado pede as derivadas parciais de
f(x,y)=\frac{\sin\bigl(e^{x^{2}y}\bigr)}{\ln(xy)}.
Para isso vamos usar a regra do quociente
\frac{\partial}{\partial x}\Bigl(\frac{N}{D}\Bigr) =\frac{N'_x\,D - N\,D'_x}{D^2},
além da regra da cadeia para derivar \sin(u) e e^{u}, e a derivada de \ln(u).

Passo 2

Calcule \frac{\partial f}{\partial x}. Defina
N(x,y)=\sin\bigl(e^{x^2y}\bigr), \quad D(x,y)=\ln(xy).
Primeiro,
N'_x =\cos\bigl(e^{x^2y}\bigr)\cdot\frac{\partial}{\partial x}\bigl(e^{x^2y}\bigr) =\cos\bigl(e^{x^2y}\bigr)\,e^{x^2y}\,\bigl(2x\,y\bigr).
Em seguida,
D'_x =\frac{1}{xy}\,\frac{\partial}{\partial x}(xy) =\frac{1}{xy}\,y =\frac{1}{x}.
Aplicando a regra do quociente,
\frac{\partial f}{\partial x} = \frac{\cos(e^{x^2y})\,e^{x^2y}\,(2xy)\,\ln(xy) -\sin(e^{x^2y})\,\frac{1}{x}} {[\ln(xy)]^2}.

Passo 3

Calcule \frac{\partial f}{\partial y} de modo análogo. Mantendo
N(x,y)=\sin\bigl(e^{x^2y}\bigr), \quad D(x,y)=\ln(xy),
temos
N'_y =\cos\bigl(e^{x^2y}\bigr)\cdot e^{x^2y}\,\frac{\partial}{\partial y}(x^2y) =\cos(e^{x^2y})\,e^{x^2y}\,(x^2),
e
D'_y =\frac{1}{xy}\,\frac{\partial}{\partial y}(xy) =\frac{1}{xy}\,x =\frac{1}{y}.
Logo,
\frac{\partial f}{\partial y} = \frac{\cos(e^{x^2y})\,e^{x^2y}\,x^2\,\ln(xy) -\sin(e^{x^2y})\,\frac{1}{y}} {[\ln(xy)]^2}.

Resposta

\frac{\partial f}{\partial x} = \frac{\cos(e^{x^2y})\,e^{x^2y}\,(2xy)\,\ln(xy) -\sin(e^{x^2y})\,\frac{1}{x}} {[\ln(xy)]^2},
\frac{\partial f}{\partial y} = \frac{\cos(e^{x^2y})\,e^{x^2y}\,x^2\,\ln(xy) -\sin(e^{x^2y})\,\frac{1}{y}} {[\ln(xy)]^2}.

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