1. Sejam
f(x, y)=e^{\frac{x}{y^{2}}}
e
X(r, s)=\cos \left(r^{2}+s^{3}\right)
e
Y(r, s)=e^{3 r-2 s}
Utilizando a regra da cadeia calcule f_{r} e f_{s}.

Passo 1

O enunciado define a função
f(x,y)=e^{\frac{x}{y^{2}}},
e as funções de mudança de variável
X(r,s)=\cos\bigl(r^{2}+s^{3}\bigr), \qquad Y(r,s)=e^{3r-2s}.
Queremos calcular as derivadas parciais
\displaystyle f_{r}=\frac{\partial}{\partial r}\bigl[f\bigl(X(r,s),Y(r,s)\bigr)\bigr]
e
\displaystyle f_{s}=\frac{\partial}{\partial s}\bigl[f\bigl(X(r,s),Y(r,s)\bigr)\bigr]
usando a regra da cadeia.

Passo 2

Calcular as derivadas parciais de f em relação a x e y:
f_{x}(x,y) =\frac{\partial}{\partial x}\,e^{\frac{x}{y^{2}}} =e^{\frac{x}{y^{2}}}\cdot\frac{1}{y^{2}},
f_{y}(x,y) =\frac{\partial}{\partial y}\,e^{\frac{x}{y^{2}}} =e^{\frac{x}{y^{2}}}\cdot\frac{\partial}{\partial y}\Bigl(\frac{x}{y^{2}}\Bigr) =e^{\frac{x}{y^{2}}}\cdot\Bigl(-\frac{2x}{y^{3}}\Bigr).

Passo 3

Calcular as derivadas de X e Y:
X_{r}(r,s) =\frac{\partial}{\partial r}\cos\bigl(r^{2}+s^{3}\bigr) =-\sin\bigl(r^{2}+s^{3}\bigr)\cdot2r,
X_{s}(r,s) =-\sin\bigl(r^{2}+s^{3}\bigr)\cdot3s^{2},
Y_{r}(r,s) =\frac{\partial}{\partial r}e^{3r-2s} =e^{3r-2s}\cdot3,
Y_{s}(r,s) =e^{3r-2s}\cdot(-2).

Passo 4

Aplicar a regra da cadeia para f_{r}:
f_{r} =f_{x}(X,Y)\,X_{r}+f_{y}(X,Y)\,Y_{r}.
Substituindo f_{x}, f_{y}, X_{r}, Y_{r} e usando X=\cos(r^{2}+s^{3}), Y=e^{3r-2s},
f_{r} =e^{\frac{X}{Y^{2}}}\Bigl(\frac{1}{Y^{2}}\bigl(-2r\sin(r^{2}+s^{3})\bigr) +\bigl(-\tfrac{2X}{Y^{3}}\bigr)\bigl(3e^{3r-2s}\bigr)\Bigr).
Como
\displaystyle -\frac{2X}{Y^{3}}\cdot3e^{3r-2s}=-\frac{6X}{Y^{2}}, obtemos
f_{r} =e^{\frac{\cos(r^{2}+s^{3})}{e^{2(3r-2s)}}}\, \Bigl(\, -\frac{2r\sin(r^{2}+s^{3})}{e^{2(3r-2s)}} -\frac{6\cos(r^{2}+s^{3})}{e^{2(3r-2s)}} \Bigr).

Passo 5

De modo análogo, para f_{s} = f_{x}X_{s}+f_{y}Y_{s}:
f_{s} =e^{\frac{X}{Y^{2}}}\Bigl(\frac{1}{Y^{2}}\bigl(-3s^{2}\sin(r^{2}+s^{3})\bigr) +\bigl(-\tfrac{2X}{Y^{3}}\bigr)\bigl(-2e^{3r-2s}\bigr)\Bigr).
Aqui
\displaystyle -\frac{2X}{Y^{3}}\cdot(-2e^{3r-2s})=\frac{4X}{Y^{2}}, logo
f_{s} =e^{\frac{\cos(r^{2}+s^{3})}{e^{2(3r-2s)}}}\, \Bigl(\, -\frac{3s^{2}\sin(r^{2}+s^{3})}{e^{2(3r-2s)}} +\frac{4\cos(r^{2}+s^{3})}{e^{2(3r-2s)}} \Bigr).

Resposta

\boxed{ f_{r}(r,s) =e^{\frac{\cos(r^{2}+s^{3})}{e^{2(3r-2s)}}} \Bigl(\, -\frac{2r\sin(r^{2}+s^{3})+6\cos(r^{2}+s^{3})}{e^{2(3r-2s)}} \Bigr), }
\boxed{ f_{s}(r,s) =e^{\frac{\cos(r^{2}+s^{3})}{e^{2(3r-2s)}}} \Bigl(\, -\frac{3s^{2}\sin(r^{2}+s^{3})-4\cos(r^{2}+s^{3})}{e^{2(3r-2s)}} \Bigr). }

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