Question

If $x=e^{y+e^{y+e^{y+\cdots\infty}}}$, where $x>0$, then $\frac{dy}{dx}$ is:

• A. $\frac{x}{1-x}$
• B. $\frac{1-x}{x}$
• C. $\frac{x-1}{x}$
• D. $\frac{x^2}{1+x}$

Answer 1

Answer: (B)

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

Answer 2

The given equation is $x=e^{y+e^{y+e^{y+\cdots\infty}}}$.

Observe the repeating pattern in the exponent. Let the entire expression on the right-hand side be $x$. The exponent part can be written as $y + (e^{y+e^{y+\cdots\infty}})$. The term in the parenthesis, $e^{y+e^{y+\cdots\infty}}$, is identical to the original expression $x$. Therefore, we can simplify the given equation as: $x = e^{y+x}$

Now, we need to find $\frac{dy}{dx}$. We can do this by first expressing $y$ in terms of $x$, or by using implicit differentiation.

Method 1: Expressing y in terms of x Take the natural logarithm on both sides of the equation $x = e^{y+x}$: $\ln(x) = \ln(e^{y+x})$ Using the property $\ln(e^A) = A$: $\ln(x) = y+x$

Now, isolate $y$: $y = \ln(x) - x$

Differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\ln(x) - x)$ Using the differentiation rules $\frac{d}{dx}(\ln u) = \frac{1}{u}$ and $\frac{d}{dx}(x) = 1$: $\frac{dy}{dx} = \frac{1}{x} - 1$

To combine the terms into a single fraction: $\frac{dy}{dx} = \frac{1}{x} - \frac{x}{x}$ $\frac{dy}{dx} = \frac{1-x}{x}$

Method 2: Implicit Differentiation Start with the simplified equation $x = e^{y+x}$. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x) = \frac{d}{dx}(e^{y+x})$ $1 = e^{y+x} \cdot \frac{d}{dx}(y+x)$ (using the chain rule, where $\frac{d}{du}(e^u) = e^u$ and $u = y+x$) $1 = e^{y+x} \cdot \left(\frac{dy}{dx} + \frac{d}{dx}(x)\right)$ $1 = e^{y+x} \cdot \left(\frac{dy}{dx} + 1\right)$

Now, substitute $x$ back for $e^{y+x}$ (from the initial simplified equation): $1 = x \left(\frac{dy}{dx} + 1\right)$

Divide both sides by $x$: $\frac{1}{x} = \frac{dy}{dx} + 1$

Isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{x} - 1$ $\frac{dy}{dx} = \frac{1-x}{x}$

Both methods yield the same result.