Answer :

GinnyAnswer
To find [tex]\(\frac{dy}{dx}\)[/tex] for the given implicit equation:
[tex]\[ 1 - \ln(xy) = e^y \][/tex]

First, we'll start by differentiating both sides of the equation implicitly with respect to [tex]\(x\)[/tex].

1. Differentiate [tex]\(1\)[/tex]:
[tex]\[ \frac{d}{dx}(1) = 0 \][/tex]

2. Differentiate [tex]\(- \ln(xy)\)[/tex]:
[tex]\[ \frac{d}{dx}[- \ln(xy)] \][/tex]
Using the chain rule, we get:
[tex]\[ -\frac{d}{dx} [\ln(xy)] = -\frac{1}{xy} \cdot \frac{d}{dx}(xy) \][/tex]
Using the product rule for differentiation:
[tex]\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y \][/tex]
So,
[tex]\[ -\frac{d}{dx} [\ln(xy)] = - \frac{1}{xy} \left(x \frac{dy}{dx} + y \right) \][/tex]

3. Differentiate [tex]\(e^y\)[/tex]:
[tex]\[ \frac{d}{dx} [e^y] = e^y \cdot \frac{dy}{dx} \][/tex]

Putting it all together, the differentiated equation is:
[tex]\[ 0 - \frac{1}{xy} (x \frac{dy}{dx} + y) = e^y \frac{dy}{dx} \][/tex]

Simplify the equation:
[tex]\[ -\frac{x \frac{dy}{dx} + y}{xy} = e^y \frac{dy}{dx} \][/tex]

[tex]\[ - \left( \frac{x \frac{dy}{dx}}{xy} + \frac{y}{xy} \right) = e^y \frac{dy}{dx} \][/tex]

[tex]\[ - \left( \frac{\frac{dy}{dx}}{y} + \frac{1}{x} \right) = e^y \frac{dy}{dx} \][/tex]

Multiply through by [tex]\(xy\)[/tex] to clear the fractions:
[tex]\[ - \left( x \frac{\frac{dy}{dx}}{y} + 1 \right) = xy e^y \frac{dy}{dx} \][/tex]

Simplify:
[tex]\[ - \left( x \frac{\frac{dy}{dx}}{y} + 1 \right) = xy e^y \frac{dy}{dx} \][/tex]

[tex]\[ - x \frac{\frac{dy}{dx}}{y} - 1 = x y e^y \frac{dy}{dx} \][/tex]

Multiply through by [tex]\(- y\)[/tex]:
[tex]\[ x \frac{dy}{dx} + y = - y x e^y \frac{dy}{dx} \][/tex]

Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ x \frac{dy}{dx} + y = - y x e^y \frac{dy}{dx} \][/tex]

Combine the [tex]\(\frac{dy}{dx}\)[/tex] terms on one side:
[tex]\[ x \frac{dy}{dx} + y = - y x e^y \frac{dy}{dx} \][/tex]

[tex]\[ x \frac{dy}{dx} + y = - y x e^y \frac{dy}{dx} \][/tex]

Factor out [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \left( x + y x e^y \right) \frac{dy}{dx} = - y \][/tex]

[tex]\(\frac{dy}{dx}\)[/tex] is alone:
[tex]\[ \frac{dy}{dx} = \frac{- y}{x + y x e^y} \][/tex]

[tex]\[ \frac{dy}{dx} = \frac{-y}{x(1 + y e^y)} \][/tex]

Thus, the simplified form of [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \boxed{\frac{y}{x(y e^y + 1)}} \][/tex]

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