If ${(\cos x)}^{y} = {(\cos y)}^{x}, find \frac{dy}{dx} .$

Solution

The given function is ( cos x )^y = ( cos y )^x

Taking logarithm on both the sides, we obtain

y log cos x = x log cos y

Differentiating both sides, we obtain

$\log \cos x \times \frac{dy}{dx} + y \times \frac{d}{dx} (\log \cos x) = \log \cos y \times \frac{d}{dx} (x) + x \times \frac{d}{dx} (\log \cos y) \Rightarrow \log \cos x \times \frac{dy}{dx} + y \times \frac{1}{\cos x} \times \frac{d}{dx} (\cos x) = \log \cos y \times 1 + x \times \frac{1}{\cos y} \times \frac{d}{dx} (\cos y) \Rightarrow \log \cos x \times \frac{dy}{dx} + \frac{y}{\cos x} (- \sin x) = \log \cos y + \frac{x}{\cos y} \times (- \sin y) \times \frac{dy}{dx}$

$\Rightarrow \log \cos x \times \frac{dy}{dx} - y \tan x = \log \cos y - x \tan y \times \frac{dy}{dx} \Rightarrow \log \cos x \times \frac{dy}{dx} + x \tan y \times \frac{dy}{dx} = \log \cos y + y \tan x \Rightarrow (\log \cos x + x \tan y) \times \frac{dy}{dx} = \log \cos y + y \tan x ∴ \frac{dy}{dx} = \frac{\log \cos y + y \tan x}{\log \cos x + x \tan y}$

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If ${(\cos x)}^{y} = {(\cos y)}^{x}, find \frac{dy}{dx} .$

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Relations And Functions

If cos x y = cos y x, find dydx.

Some More Questions From Relations and Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

If ${(\cos x)}^{y} = {(\cos y)}^{x}, find \frac{dy}{dx} .$