Question

If $y = \sqrt{\log x + \sqrt{\log x + \sqrt{\log x + ....\infty }}}$, show that $\frac{dy}{dx} = \frac{1}{x(2y-1)}$

Answer 1

hey mate
here's the solution

Answer 2

Answer:

$\dfrac{dy}{dx} = \dfrac {1}{x(2y - 1)}$

Step-by-step explanation:

Given:

$y = \sqrt{logx + {\sqrt{logx+ \sqrt{logx + \sqrt{logx + ...... \infty}}}}}\\\\y = \sqrt{logx + y} \\\\y^2 = logx + y........ (1)$

Now differentiate the equation (1)

$2y\dfrac{dy}{dx}= \dfrac 1x + \dfrac{dy}{dx} \\\\ 2y\dfrac{dy}{dx} - \dfrac{dy}{dx}= \dfrac 1x\\\\\dfrac{dy}{dx}(2y - 1)= \dfrac 1x\\\\\dfrac{dy}{dx} = \dfrac {1}{x(2y - 1)}$

$\dfrac{dy}{dx} = \dfrac {1}{x(2y - 1)}$

Hence Proved.