Practicing Success
If $(x+1) e^y=1$, then : |
$\frac{d^2 y}{d x^2}=y^2$ $\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2$ $\frac{d^2 y}{d x^2}=-y^2$ $\frac{d^2 y}{d x^2}=\frac{d y}{d x}$ |
$\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2$ |
$(x+1) e^y=1$ so $e^y=\frac{1}{1+x}$ ........(1) differentiating both sides wrt x $e^y \frac{d y}{d x}=\frac{-1}{(1+x)^2} \Rightarrow \frac{d y}{d x}=\frac{-1}{(1+x)^2 e y}$ $\Rightarrow \frac{d y}{d x}=-\frac{(1+x)}{(1+x)^2}$ from (1) $\Rightarrow \frac{d y}{d x}=\frac{-1}{(1+x)}$ again differentiating wrt x $\frac{d y}{d x}=\frac{-1}{1+x} \Rightarrow\left(\frac{d y}{d x}\right)^2=\frac{1}{(1+x)^2}$ $\rightarrow \frac{d^2 y}{d x^2} =\frac{1}{(1+x)^2}$ $\frac{d^2 y}{d x^2} =\left(\frac{d y}{d x}\right)^2$ |