Question

If $x={\log }_{e}t,t>0$ and $y+1=t^2$. Then $\frac{d^{2}y}{d{x}^2}$ is equal to:

• A. $4e^{2x}$
• B. $-\frac{1}{2}{e}^{-4x}$
• C. $-\frac{3}{4}{e}^{-5x}$
• D. $4e^x$

Answer 1

The correct option is B

$-\frac{1}{2}{e}^{-4x}$

Explanation for the correct option.

$$\begin{gathered} x={\log }_{e}t \\ \Rightarrow e^x=t.......\left(1\right) \end{gathered}$$

$\begin{array}{rcl}y+1& =& t^2\\ & \Rightarrow & y+1={\left(e^x\right)}^2\\ & \Rightarrow & y+1=e^{2x}\end{array}$

By differentiating w.r.t $y$, we get

$\begin{array}{rcl}1+0& =& 2e^{2x}\frac{dx}{dy}\\ & \Rightarrow & \frac{dx}{dy}=\frac{1}{2}{e}^{-2x}\end{array}$

Now,

$\begin{array}{rcl}\frac{d^{2}x}{d{y}^2}& =& \frac{1}{2}\left(-2\right)\left(e^{-2x}\right)\frac{dx}{dx}\\ & =& -e^{-2x}\left(\frac{1}{2}{e}^{-2x}\right)\\ & =& -\frac{1}{2}{e}^{-4x}\end{array}$

Thus, option B is correct.