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