Question:medium

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

Show 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.
Updated On: Mar 30, 2026
  • $\frac{x^2+1}{2y-1}$
  • $\frac{2x}{2y-1}$
  • $\frac{1}{(x^2+1)(2y-1)}$
  • $\frac{2x}{(x^2+1)(2y-1)}$
Show Solution

The Correct Option is D

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