Q16  If  e ^y (x+1) = 1  show that \frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2
 

Answers (1)

Given function is
e ^y (x+1) = 1
We can rewrite it as
e^y = \frac{1}{x+1}
Now, differentiation w.r.t. x
\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}                 -(i)
Now, second order derivative is 
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}        -(ii)
Now, we need to show
\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2
Put value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}  from equation (i) and (ii)
 \frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2
                    =\frac{1}{(x+1)^2}
Hence, L.H.S. = R.H.S.
Hence proved
 

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