Question

If $x=e^{y+e^{y+e^{y+...}}}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{x}$
• B. $\frac{1-x}{x}$
• C. $\frac{x}{1+x}$
• D. None of these

Answer 1

The correct option is B

$\frac{1-x}{x}$

Explanation for the correct option.

Step 1. Simplify the given equation.

The exponent in the right hand side starts repeating itself in the given equation $x=e^{y+e^{y+e^{y+...}}}$. It can be written as:

$$\begin{gathered} x=e^{y+e^{y+e^{y+...}}} \\ \Rightarrow x=e^{y+x} \end{gathered}$$

Now, take natural logarithm both sides:

$$\begin{gathered} \ln \left(x\right)=\ln \left(e^{y+x}\right) \\ \Rightarrow \ln \left(x\right)=\left(y+x\right)\ln \left(e\right) \\ \Rightarrow \ln \left(x\right)=y+x \end{gathered}$$

Step 2. Find the value of $\frac{dy}{dx}$.

Differentiate both sides of the equation $\ln \left(x\right)=y+x$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(\ln \right(x\left)\right)=\frac{d}{dx}(y+x) \\ \Rightarrow \frac{1}{x}=\frac{dy}{dx}+1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x}-1 \\ \Rightarrow \frac{dy}{dx}=\frac{1-x}{x} \end{gathered}$$

Hence, the correct option is B.