Question

If $e^y(x+1)=1,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\frac{d^{2}y}{{\mathrm{dx}}^2}$ is equal to

• A. ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2$
• B. $\frac{1}{{\left(x+1\right)}^2}$
• C. $-\frac{1}{{\left(x+1\right)}^2}$
• D. $\frac{1}{4}\mathrm{at}x=1$

Answer 1

Answer: (A), (B), (D)

$\frac{1}{4}\mathrm{at}x=1$

We have,
$e^y\left(x+1\right)=1\Rightarrow e^y=\frac{1}{x+1}\mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\Rightarrow y=\log \left(\frac{1}{x+1}\right)\Rightarrow y=-\log (x+1)\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{x+1}\mathrm{Differentiating}\mathrm{again}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\frac{d}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{d}{\mathrm{dx}}\left(-\frac{1}{x+1}\right)\Rightarrow \frac{d^{2}y}{{\mathrm{dx}}^2}=\frac{1}{{\left(x+1\right)}^2}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\mathrm{At}x=1,{\left(\frac{d^{2}y}{{\mathrm{dx}}^2}\right)}_{x=1}=\frac{1}{{\left(1+1\right)}^2}=\frac{1}{4}$