#### Question

If \[e^{x + y} - x = 0\] ,prove that \[\frac{dy}{dx} = \frac{1 - x}{x}\] ?

#### Solution

\[\text{ We have}, e^{x + y} - x = 0\]

\[ \Rightarrow e^{x + y} = x . . . \left( 1 \right)\]

Differentiating with respect to* x* using chain rule,

\[\frac{d}{dx}\left( e^{x + y} \right) = \frac{d}{dx}\left( x \right)\]

\[ \Rightarrow e^{x + y} \frac{d}{dx}\left( x + y \right) = 1\]

\[ \Rightarrow x\left[ 1 + \frac{dy}{dx} \right] = 1 \left[ \text{ Using equation } \left( i \right) \right]\]

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{1}{x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x} - 1\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1 - x}{x}\]

Is there an error in this question or solution?

Solution If E X + Y − X = 0 ,Prove that D Y D X = 1 − X X ? Concept: Simple Problems on Applications of Derivatives.