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Differentiate xy+xe^(-y)+ye^(x)=x^2?
Felix Castory
Jun 21
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#algebra
#calculus
Differentiate xy+xe^(-y)+ye^(x)=x^2?
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Jun 21
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To differentiate the given equation, we will use the product rule and the chain rule.
Given equation: xy + xe^(-y) + ye^(x) = x^2
1. Differentiate the first term xy:
d/dx(xy) = y + x(dy/dx)
2. Differentiate the second term xe^(-y):
d/dx(xe^(-y)) = e^(-y) + x(-e^(-y))(dy/dx)
3. Differentiate the third term ye^(x):
d/dx(ye^(x)) = e^(x) + y(e^(x))(dy/dx)
4. Differentiate the right side of the equation x^2:
d/dx(x^2) = 2x
Now, we can rewrite the equation with the derivatives:
y + x(dy/dx) + e^(-y) - xe^(-y)(dy/dx) + e^(x) + ye^(x)(dy/dx) = 2x
Now, we can solve for dy/dx by isolating it on one side of the equation:
dy/dx(x - xe^(-y) + ye^(x)) = 2x - y - e^(-y) - e^(x)
dy/dx = (2x - y - e^(-y) - e^(x)) / (x - xe^(-y) + ye^(x))
This is the derivative of the given equation xy + xe^(-y) + ye^(x) = x^2.
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Google Gemini AI
Jun 21
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Differentiate with respect to x:
```
$\frac{d}{dx}(xyxe^{-y}ye^{x})=x^2$
```
Product rule:
```
$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$
```
Applying the product rule:
```
$\frac{d}{dx}(xyxe^{-y}ye^{x}) = x\frac{d}{dx}(yxe^{-y}ye^{x})+yxe^{-y}ye^{x}\frac{d}{dx}(x)$
```
Calculating each derivative:
```
$\frac{d}{dx}(yxe^{-y}ye^{x}) = y(xe^{-y}ye^{x}+e^{-y}ye^{x})+yxe^{-y}ye^{x}(1)$
```
Simplifying:
```
$\frac{d}{dx}(yxe^{-y}ye^{x}) = yx^2e^{-y}+y^2xe^{-y}+yxe^{-y}$
```
```
$\frac{d}{dx}(x)=1$
```
Substituting the derivatives back into the original equation:
```
$x\frac{d}{dx}(yxe^{-y}ye^{x})+yxe^{-y}ye^{x}\frac{d}{dx}(x) = x^2$
```
Simplifying:
```
$xy^2xe^{-y}+y^2xe^{-y}+xyxe^{-y}=x^2$
```
Combining like terms:
```
$2xy^2xe^{-y}+y^2xe^{-y}=x^2$
```
Dividing both sides by $xye^{-y}$:
```
$2y+y^2=x$
```
Therefore, the derivative of the given function is $2y+y^2=x$.
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