Question

If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\cdots \infty }}}$, then prove that $\dfrac{dy}{dx}=\dfrac{1}{x\left( 2y-1 \right)}$

Answer 1

Assume that $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\cdots \infty }}}$. Square both sides and hence prove that ${{y}^{2}}=\log x+y$
Differentiate both sides of the equation and using the method of implicit differentiation and hence prove that $\dfrac{dy}{dx}=\dfrac{1}{x\left( 2y-1 \right)}$

Complete step-by-step answer:
Before solving the question, we need to know how to deal with infinite sequences, like infinite radicals, sum to infinity, infinite continued product etc.
Dealing with infinite sequence:
When trying to solve the questions involving infinite sequences, we try to express the result in terms of itself and solve the resulting equation.
In case radicals e.g. $\sqrt{x+\sqrt{x+\sqrt{x+\cdots \infty }}}$, we substitute it for y and square on both sides.
In case of sums e.g. $a+ar+a{{r}^{2}}+\cdots$ we substitute it for y and observe that $y=a+r\left( a+ar+a{{r}^{2}}+\cdots \right)=a+ry$ and solve the resulting expression.
So, let $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\cdots \infty }}}$
Squaring both sides, we get
${{y}^{2}}=\log x+\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\cdots \infty }}}$
Hence, we have
${{y}^{2}}=\log x+y$
Subtracting $\log x+y$ from both sides, we get
${{y}^{2}}-\log x-y=0$
Differentiating both sides, with respect to y, we get
$\dfrac{d}{dx}\left( {{y}^{2}} \right)-\dfrac{d}{dx}\left( \log x \right)-\dfrac{d}{dx}\left( y \right)=0$
Now, we have
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=\dfrac{d}{dg\left( x \right)}f\left( g\left( x \right) \right)\times \dfrac{dg\left( x \right)}{dx}$. This is known as chain rule of differentiation.
Hence, we have
$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}=2y\dfrac{dy}{dx}$
Also, we know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
Hence, we have
$2y\dfrac{dy}{dx}-\dfrac{1}{x}-\dfrac{dy}{dx}=0$
Taking $\dfrac{dy}{dx}$ common from the first and the last term on LHS, we get
$\left( 2y-1 \right)\dfrac{dy}{dx}-\dfrac{1}{x}=0$
Adding $\dfrac{1}{x}$ on both sides, we get
$\left( 2y-1 \right)\dfrac{dy}{dx}=\dfrac{1}{x}$
Dividing both sides by 2y-1, we get
$\dfrac{dy}{dx}=\dfrac{1}{\left( 2y-1 \right)x}$
Hence, we have
$\dfrac{dy}{dx}=\dfrac{1}{x\left( 2y-1 \right)}$
Q.E.D

Note: [1] In these types of questions, we always express y as f(x,y) and use a method of implicit differentiation to find the derivative of y with respect to x. Although in the above question, it is possible to explicitly write y in terms of x using the quadratic formula, it has been avoided as it would make the differentiation cumbersome.