Question

Differentiate $y=e^{{x^2+e}^{{x^2+e}^{x^2+e}....\infty }}$

• A. $\frac{dy}{dx}=\frac{y}{1-y}$
• B. $\frac{dy}{dx}=\frac{1}{1-y}$
• C. $\frac{dy}{dx}=\frac{2xy}{1-y}$
• D. None of these

Answer 1

The correct option is C $\frac{dy}{dx}=\frac{2xy}{1-y}$

As the function keeps on repeating to infinity we can write the terms leaving the first term as y and it would not make any difference so we have,

$\Rightarrow y=e^{x^2+y}$

Taking \ln on both sides we get,

$\Rightarrow \ln y=x^2+y$

Differentiating with respect to $x$ we get,

$\Rightarrow \frac{1}{y}.\frac{dy}{dx}=2x+\frac{dy}{dx}$

$\Rightarrow (\frac{1}{y}-1).\frac{dy}{dx}=2x$

$\Rightarrow \left(\frac{1-y}{y}\right).\frac{dy}{dx}=2x$

$\Rightarrow \frac{dy}{dx}=\frac{2xy}{1-y}$ .....Answer