Question

If $y = \sqrt{\log(x^2+1)+\sqrt{\log(x^2+1)+\sqrt{\log(x^2+1)+...}}}$, $|x|<1$, then $\frac{dy}{dx} =$

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

Hint:

For functions defined by infinite nested expressions like $y = \sqrt{f(x)+\sqrt{f(x)+...}}$, you can write them as a simple equation, $y = \sqrt{f(x)+y}$. Squaring both sides gives $y^2 = f(x)+y$, which can then be easily differentiated using implicit differentiation.

Answer 1

The Correct Option is D

The given equation involves an infinite nested radical. We can write it as:
$y = \sqrt{\log(x^2+1) + y}$.
To remove the radical, we square both sides of the equation:
$y^2 = \log(x^2+1) + y$.
Rearrange the terms to prepare for implicit differentiation:
$y^2 - y = \log(x^2+1)$.
Now, we differentiate both sides with respect to $x$:
$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\log(x^2+1))$.
Using the chain rule on both sides:
$2y\frac{dy}{dx} - 1\frac{dy}{dx} = \frac{1}{x^2+1} \cdot \frac{d}{dx}(x^2+1)$.
Factor out $\frac{dy}{dx}$ on the left side:
$(2y-1)\frac{dy}{dx} = \frac{1}{x^2+1} \cdot (2x)$.
Finally, solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{2x}{(x^2+1)(2y-1)}$.