R S Aggarwal Solutions for Exercise 7: EXERCISE

If  $y={\left(\sin x\right)}^{{\left(\sin x\right)}^{\left(\sin x\right)...\infty }}$, prove that $\frac{dy}{dx}=\frac{y^2\cot x}{(1-y \log \sin x)}$.

If $y=(\cos x{)}^{(\cos x{)}^{\left(\cos x\right)...\infty }}$, prove that $\frac{dy}{dx}=\frac{-y^2\tan x}{\left(1-y\log \cos x\right)}$

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

If $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x}+\dots \infty }}$, prove that$\frac{dy}{dx}=\frac{\sin x}{(1-2y)}$.

If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\dots \infty }}}$ prove that $\frac{dy}{dx}=\frac{{\sec }^{2}x}{(2y-1)}$

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

If $y=a^{x^{a^{x...\infty }}}$, prove that $\frac{dy}{dx}=\frac{y^2(\log y)}{x[1-y(\log x\left)\right(\log y\left)\right]}$

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