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]
