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  • If y=(cosx) , show that dy/dx = y^2tanx

If \mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x}) \ldots . \infty}}Show that  \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}^{2} \tan \mathrm{x}}{\mathrm{y} \log \cos \mathrm{x}-1}

Answers (1)

Given: \mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x}) \ldots . \infty}}

⇒ y = (cos x)y

Taking log both the sides, we get

log y = y log(cos x)

On differentiating both the sides, we get

\\ \frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x} \log (\cos x)+\log \cos x \frac{d}{d x} y \\ \Rightarrow \frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{\cos x} \times(-\sin x)+\log \cos x \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x}\left[\frac{1}{y}-\log \cos x\right]=-y \tan x \because\left[\frac{\sin x}{\cos x}=\tan x\right] \\ \Rightarrow \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log \cos x} \\ \Rightarrow \frac{d y}{d x}=\frac{y \tan x}{\log \cos x-\frac{1}{y}} \\ \Rightarrow \frac{d y}{d x}=\frac{y^{2} \tan x}{y \log \cos x-1}

Hence Proved

Posted by

infoexpert22

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