Calcule a derivada parcial de primeira ordem:
C(x, y)=\frac{200}{x^{2}}\left[e^{-0,02 \frac{(y-10)^{2}}{x^{2}}}+e^{-0,02 \frac{(y+10)^{2}}{x^{2}}}\right]

Passo 1

O enunciado pede as derivadas parciais de primeira ordem de
C(x,y)=\frac{200}{x^{2}}\Bigl[e^{-0,02\frac{(y-10)^{2}}{x^{2}}}+e^{-0,02\frac{(y+10)^{2}}{x^{2}}}\Bigr].
Vamos usar a regra do produto e a regra da cadeia para calcular \frac{\partial C}{\partial x} e \frac{\partial C}{\partial y}.

Passo 2

Para \frac{\partial C}{\partial x} escrevemos
C(x,y)=200\,x^{-2}\bigl(e_{1}+e_{2}\bigr),
com
e_{1}=e^{-0,02\frac{(y-10)^{2}}{x^{2}}} \quad\text{e}\quad e_{2}=e^{-0,02\frac{(y+10)^{2}}{x^{2}}}.
Pela regra do produto:
\frac{\partial C}{\partial x} =200\bigl(-2x^{-3}\bigr)(e_{1}+e_{2}) +200\,x^{-2}\Bigl(\frac{\partial e_{1}}{\partial x}+\frac{\partial e_{2}}{\partial x}\Bigr).
Agora, para cada i=1,2, defina
u_{i}=-0,02\frac{(y\pm10)^{2}}{x^{2}},
onde o sinal + vale para i=2 e o para i=1.
Então
\frac{du_{i}}{dx} =-0,02(y\pm10)^{2}\,(-2)x^{-3} =0,04(y\pm10)^{2}x^{-3},
\frac{\partial e_{i}}{\partial x} =e_{i}\,\frac{du_{i}}{dx} =0,04(y\pm10)^{2}x^{-3}e_{i}.

Passo 3

Substituindo em \partial C/\partial x e agrupando:
\frac{\partial C}{\partial x} =200\Bigl[-2x^{-3}(e_{1}+e_{2}) +x^{-2}\bigl(0,04(y-10)^{2}x^{-3}e_{1}+0,04(y+10)^{2}x^{-3}e_{2}\bigr)\Bigr]
=\frac{200}{x^{5}}\Bigl[-2x^{2}(e_{1}+e_{2}) +0,04\bigl((y-10)^{2}e_{1}+(y+10)^{2}e_{2}\bigr)\Bigr].

Passo 4

Para \frac{\partial C}{\partial y} só há derivada nos expoentes. Ainda com u_{i} acima,
\frac{du_{i}}{dy} =-0,02\cdot2(y\pm10)x^{-2} =-0,04(y\pm10)x^{-2},
\frac{\partial e_{i}}{\partial y} =e_{i}\,\frac{du_{i}}{dy} =-0,04(y\pm10)x^{-2}e_{i}.
Como 200\,x^{-2} é constante em y:
\frac{\partial C}{\partial y} =200\,x^{-2}\bigl(\tfrac{\partial e_{1}}{\partial y}+\tfrac{\partial e_{2}}{\partial y}\bigr) =-200\,x^{-2}\bigl[0,04(y-10)x^{-2}e_{1}+0,04(y+10)x^{-2}e_{2}\bigr].

Passo 5

Agrupando fatores em \partial C/\partial y:
\frac{\partial C}{\partial y} =-\frac{8}{x^{4}}\Bigl[(y-10)e^{-0,02\frac{(y-10)^{2}}{x^{2}}} +(y+10)e^{-0,02\frac{(y+10)^{2}}{x^{2}}}\Bigr].

Resposta

\frac{\partial C}{\partial x} =\frac{200}{x^{5}}\Bigl[-2x^{2}\bigl(e^{-0,02\frac{(y-10)^{2}}{x^{2}}} +e^{-0,02\frac{(y+10)^{2}}{x^{2}}}\bigr) +0,04\bigl((y-10)^{2}e^{-0,02\frac{(y-10)^{2}}{x^{2}}} +(y+10)^{2}e^{-0,02\frac{(y+10)^{2}}{x^{2}}}\bigr)\Bigr],
\frac{\partial C}{\partial y} =-\frac{8}{x^{4}}\Bigl[(y-10)e^{-0,02\frac{(y-10)^{2}}{x^{2}}} +(y+10)e^{-0,02\frac{(y+10)^{2}}{x^{2}}}\Bigr].

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