Question

Given if ${{\left( \cos x \right)}^{y}}={{\left( \cos y \right)}^{x}}$, then find the value of $\dfrac{dy}{dx}$?

Answer 1

We start solving the problem by applying logarithms to the given function ${{\left( \cos x \right)}^{y}}={{\left( \cos y \right)}^{x}}$. We differentiate after applying the logarithms using the uv rule of differentiation. We differentiate and make adjustments to get the coefficients of $\dfrac{dy}{dx}$ on one side. We make necessary calculations to get the required result.

Complete step by step answer:
Given that we have ${{\left( \cos x \right)}^{y}}={{\left( \cos y \right)}^{x}}$. We need to find the value of $\dfrac{dy}{dx}$.
We have got the value ${{\left( \cos x \right)}^{y}}={{\left( \cos y \right)}^{x}}$ ---(1).
Applying log on both sides in equation (1).
We have got the value ${{\log }_{e}}{{\left( \cos x \right)}^{y}}={{\log }_{e}}{{\left( \cos y \right)}^{x}}$ ---(2).
We know that ${{\log }_{a}}\left( {{x}^{y}} \right)=y{{\log }_{a}}\left( x \right)$. We use this result in equation (2).
We have got the value $y.{{\log }_{e}}\left( \cos x \right)=x.{{\log }_{e}}\left( \cos y \right)$.
We apply differentiation with respect to x on both sides.
We have got the value $\dfrac{d}{dx}\left( y.{{\log }_{e}}\left( \cos x \right) \right)=\dfrac{d}{dx}\left( x.{{\log }_{e}}\left( \cos y \right) \right)$ ---(3).
We know that the differentiation of the function uv is defined as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. We use this result in equation (3).
We have got the value $y.\dfrac{d{{\log }_{e}}\left( \cos x \right)}{dx}+\dfrac{dy}{dx}.{{\log }_{e}}\left( \cos x \right)=x.\dfrac{d{{\log }_{e}}\left( \cos y \right)}{dx}+\dfrac{dx}{dx}.{{\log }_{e}}\left( \cos y \right)$ ---(4).
We know that $\dfrac{d\left( \log \left( f\left( x \right) \right) \right)}{dx}=\dfrac{1}{f\left( x \right)}.\dfrac{d\left( f\left( x \right) \right)}{dx}$, and $\dfrac{dx}{dx}=1$. We use this results in equation (4).
We have got the value $y.\dfrac{1}{\cos x}.\dfrac{d\left( \cos x \right)}{dx}+\dfrac{dy}{dx}.{{\log }_{e}}\left( \cos x \right)=x.\dfrac{1}{\cos y}.\dfrac{d\left( \cos y \right)}{dx}+1.{{\log }_{e}}\left( \cos y \right)$ ---(5).
We know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$ and $\dfrac{d\left( \cos \left( f\left( x \right) \right) \right)}{dx}=-\sin \left( f\left( x \right) \right).\dfrac{d\left( f\left( x \right) \right)}{dx}$. We use this results in (5).
We have got the value $y.\dfrac{1}{\cos x}.\left( -\sin x \right)+\dfrac{dy}{dx}.{{\log }_{e}}\left( \cos x \right)=x.\dfrac{1}{\cos y}.\left( -\sin y.\dfrac{dy}{dx} \right)+1.{{\log }_{e}}\left( \cos y \right)$.
We have got the value $\dfrac{-y\sin x}{\cos x}+\dfrac{dy}{dx}.{{\log }_{e}}\left( \cos x \right)=\dfrac{-x\sin y}{\cos y}.\left( \dfrac{dy}{dx} \right)+{{\log }_{e}}\left( \cos y \right)$.
We have got the value $\dfrac{dy}{dx}.{{\log }_{e}}\left( \cos x \right)+\dfrac{x\sin y}{\cos y}.\left( \dfrac{dy}{dx} \right)=\dfrac{y\sin x}{\cos x}+{{\log }_{e}}\left( \cos y \right)$ ---(6).
We know that $\dfrac{\sin x}{\cos x}=\tan x$. We use this result in equation (6).
We have got the value $\dfrac{dy}{dx}\times \left( {{\log }_{e}}\left( \cos x \right)+x\tan y \right)=y\tan x+{{\log }_{e}}\left( \cos y \right)$.
We have got the value $\dfrac{dy}{dx}=\dfrac{y\tan x+{{\log }_{e}}\left( \cos y \right)}{{{\log }_{e}}\left( \cos x \right)+x\tan y}$.
We have got the value $\dfrac{dy}{dx}=\dfrac{y\tan x+{{\log }_{e}}\left( \cos y \right)}{x\tan y+{{\log }_{e}}\left( \cos x \right)}$.
We have found the value for $\dfrac{dy}{dx}$ as $\dfrac{y\tan x+{{\log }_{e}}\left( \cos y \right)}{x\tan y+{{\log }_{e}}\left( \cos x \right)}$.

∴ The result for $\dfrac{dy}{dx}$ is $\dfrac{y\tan x+{{\log }_{e}}\left( \cos y \right)}{x\tan y+{{\log }_{e}}\left( \cos x \right)}$.

Note: Whenever we have problems containing functions which cannot be disintegrated to single variables, we apply logarithm on both sides. We should not make any sign mistakes while making the differentiation. Similarly, we get the problems to solve the problem by having powers as trigonometric functions.