Question

DIFFERENTIAL CALCULUS
If e^(x+y)=x-y, show that
dy/dx=y(1-x) / x(y-1)


please help out with steps XD

Answer 1

Hope it will help you .

Answer 2

Answer:

Proved.

Step-by-step explanation:

Given:- $e^{x+y} = xy$

To Prove:- $\frac{dy}{dx} = \frac{y(1 - x}{x(y - 1)}$

Proof:-

The given expression of Differential calculus is $e^{x+y} = xy$.

Applying log on both the sides of the above equation, we get

⇒ log($e^{x+y}$) = log( xy )

⇒ (x + y) log e = log( xy )

⇒ (x + y) = log( xy )     [ ∵ log e = 1 ]

Now differentiate y with respect to x on both sides.

⇒ $\frac{d}{dx} (x+y) = \frac{d}{dx}[log(xy)]$

⇒ 1 + $\frac{dy}{dx}$ = $\frac{1}{xy} (x \frac{dy}{dx} + y)$  [ ∵ derivative of log x = 1/x.]

⇒ 1 +  $\frac{dy}{dx}$ = $\frac{1}{y}\frac{dy}{dx} + \frac{1}{x}$

⇒ $\frac{dy}{dx} - \frac{1}{y}\frac{dy}{dx}$ = $\frac{1}{x}$ - 1

⇒ $\frac{dy}{dx}(1-\frac{1}{y} )$ = $\frac{1-x}{x}$

⇒ $\frac{dy}{dx} (\frac{y-1}{y} )$ = $\frac{1-x}{x}$

⇒ $\frac{dy}{dx}$ = $(\frac{1-x}{x})(\frac{y}{y-1} )$

⇒ $\frac{dy}{dx}$ = $\frac{y(1-x)}{x(y-1)}$

Hence proved when $e^{x+y}$ = x -y then $\frac{dy}{dx}$ = $\frac{y(1-x)}{x(y-1)}$ .

#SPJ3

To understand what is differential calculus, visit here

https://brainly.in/question/4027846

Know the difference between differential calculus and integral calculus here

https://brainly.in/question/14097456