Question

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

Answer 1

here is your answer...

Answer 2

$\dfrac{dy}{dx} =-e^{y},$ proved.

Step-by-step explanation:

We have,

$e^{y} (x+1)=1$        ...... (1)

To prove that, $\dfrac{dy}{dx} =-e^{y}$

We know that,

$\dfrac{d(uv)}{dx} =u\dfrac{dv}{dx} +v\dfrac{du}{dx}$

Differentiating (1) w.r.t. x, we get

$\dfrac{d(e^{y} (x+1))}{dx} =\dfrac{d(1)}{dx}$

⇒ $e^{y} (1+0)+(x+1)e^{y}\dfrac{dy}{dx} =0$

⇒$e^{y} =-(x+1)e^{y}\dfrac{dy}{dx}$

⇒$1=-(x+1)\dfrac{dy}{dx}$

⇒$\dfrac{dy}{dx}=-\dfrac{1}{x+1}$

Using (1), we get

⇒ $\dfrac{dy}{dx} =-e^{y}$, proved.

Hence, $\dfrac{dy}{dx} =-e^{y}$, proved.