Question

If $e^y(x+1)=1$, show that $y_2=y_1^2$.

Answer 1

$\begin{array}{c}(x+1)=e^{-y}\\ differentiatewithrespecttox\\ \Rightarrow \frac{d(x+1)}{dx}=\frac{d\left(e^{-y}\right)}{dx}\\ 1=-e^{-y}\times \left(\frac{dy}{dx}\right)\\ -ey=\left(\frac{dy}{dx}\right).......\left(i\right)\\ Thenagaindifferentiatew.r.tox\\ \frac{-d\left(e^y\right)}{dx}=\frac{d\left(\frac{dy}{dx}\right)}{dx}\\ \Rightarrow -e^y.\frac{dy}{dx}=\frac{d^{2}y}{dx}\\ \Rightarrow \left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)=\frac{d^{2}y}{d{x}^2}\\ \left[\because y_2=\frac{d^{2}y}{d{x}^2}\right]\\ \left[\because y_1=\frac{dy}{dx}\right]\\ {y_1}^2-=y_2\end{array}$