Question

If $y=(\cos x{)}^{(\cos x{)}^{(\cos x{)}^{{.}^{{.}^{{.}^{\infty }}}}}}$, then $\frac{dy}{dx}$ is equal to

• A. $-\frac{y^2\tan x}{1-yln\left(\cos x\right)}$
• B. $\frac{y^2\tan x}{1+yln\left(\cos x\right)}$
• C. $\frac{y^2\tan x}{1-yln\left(\sin x\right)}$
• D. $\frac{y^2\sin x}{1+yln\left(\sin x\right)}$

Answer 1

The correct option is A $-\frac{y^2\tan x}{1-yln\left(\cos x\right)}$

$y={\left(cos\right)}^y$

Taking log both sides, $log\left(y\right)=ylog\left(cosx\right)$

Differentiating with respect to x,

$\frac{1}{y}\frac{dy}{dx}=y\frac{(-sinx)}{cosx}+\frac{dy}{dx}log\left(cosx\right)$

$\frac{1}{y}\frac{dy}{dx}(1-ylog\left(cosx\right))=\frac{-ysinx}{cosx}=-ytanx$

Or, $\frac{dy}{dx}=\frac{-y^{2}tanx}{1-ylog\left(cosx\right)}$