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Question: If function is given by \({{x}^{y}}={{e}^{x-y}}\) then \(\dfrac{dy}{dx}=\) (a) \(\dfrac{\log x}{{...

If function is given by xy=ex−y{{x}^{y}}={{e}^{x-y}} then dydx=\dfrac{dy}{dx}=
(a) log⁡x(1+log⁡x)2\dfrac{\log x}{{{\left( 1+\log x \right)}^{2}}}
(b) −log⁡x(1+log⁡x)2-\dfrac{\log x}{{{\left( 1+\log x \right)}^{2}}}
(c) −log⁡x(1−log⁡x)2-\dfrac{\log x}{{{\left( 1-\log x \right)}^{2}}}
(d) log⁡x(1−log⁡x)2\dfrac{\log x}{{{\left( 1-\log x \right)}^{2}}}

Explanation

Solution

Here, first we will convert the given equation xy=ex−y{{x}^{y}}={{e}^{x-y}} in form of log by using the formula nlog⁡m=log⁡mnn\log m=\log {{m}^{n}} . Then, we will have an equation like this ylog⁡x=(x−y)y\log x=\left( x-y \right) . We will then differentiate this with respect to x. We will be using the formula ddx(xy)=xdydx+dxdx⋅y\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+\dfrac{dx}{dx}\cdot y , ddxlog⁡x=1x\dfrac{d}{dx}\log x=\dfrac{1}{x} , ddxx=1\dfrac{d}{dx}x=1 . Thus, on solving we will get the answer.

Complete step-by-step solution:
Here, we are given the equation xy=ex−y{{x}^{y}}={{e}^{x-y}} . So, we will use the formula nlog⁡m=log⁡mnn\log m=\log {{m}^{n}} .
We will first multiply with log⁡\log on both the sides. So, we get as
log⁡xy=log⁡ex−y\log {{x}^{y}}=\log {{e}^{x-y}}
Now, we will apply the formula nlog⁡m=log⁡mnn\log m=\log {{m}^{n}} so, we get as
ylog⁡x=(x−y)log⁡ey\log x=\left( x-y \right)\log e
Now, we will assume that log⁡e=log⁡ee=1\log e={{\log }_{e}}e=1 because we know the formula log⁡nn=1{{\log }_{n}}n=1 . So, we get equation as
ylog⁡x=(x−y)y\log x=\left( x-y \right) ……………..(1)
Now, we will differentiate the equation with respect to x. We can write it as
ddx(ylog⁡x)=ddx(x−y)\dfrac{d}{dx}\left( y\log x \right)=\dfrac{d}{dx}\left( x-y \right)
Here, we will use the product rule i.e. given as ddx(xy)=xdydx+dxdx⋅y\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+\dfrac{dx}{dx}\cdot y . So, here we will take x as y and y as log⁡x\log x . So, we can write it as
yddxlog⁡x+dydx⋅log⁡x=ddxx−dydxy\dfrac{d}{dx}\log x+\dfrac{dy}{dx}\cdot \log x=\dfrac{d}{dx}x-\dfrac{dy}{dx}
Now, we know that differentiation of ddxlog⁡x=1x\dfrac{d}{dx}\log x=\dfrac{1}{x} , ddxx=1\dfrac{d}{dx}x=1 . So, we can write it as
y⋅1x+dydx⋅log⁡x=1−dydxy\cdot \dfrac{1}{x}+\dfrac{dy}{dx}\cdot \log x=1-\dfrac{dy}{dx}
Now, taking dydx\dfrac{dy}{dx} terms on left side and other terms on right side, we get as
dydx+dydx⋅log⁡x=1−y⋅1x\dfrac{dy}{dx}+\dfrac{dy}{dx}\cdot \log x=1-y\cdot \dfrac{1}{x}
On taking dydx\dfrac{dy}{dx} common, we get as
dydx(1+log⁡x)=1−yx\dfrac{dy}{dx}\left( 1+\log x \right)=1-\dfrac{y}{x}
On further simplification, we can write it as
dydx=x−yx(1+log⁡x)\dfrac{dy}{dx}=\dfrac{x-y}{x\left( 1+\log x \right)}
Now, from equation (1), we will put value of x−yx-y so, we get as
dydx=ylog⁡xx(1+log⁡x)\dfrac{dy}{dx}=\dfrac{y\log x}{x\left( 1+\log x \right)} ………………..(2)
Now, from equation (1) i.e. ylog⁡x=(x−y)y\log x=\left( x-y \right) we can write it as ylog⁡x+y=xy\log x+y=x . On taking y common we get as
y(log⁡x+1)=xy\left( \log x+1 \right)=x
yx=1(log⁡x+1)\dfrac{y}{x}=\dfrac{1}{\left( \log x+1 \right)}
We will substitute this value in the equation (2) so, we get as
dydx=log⁡x(1+log⁡x)2\dfrac{dy}{dx}=\dfrac{\log x}{{{\left( 1+\log x \right)}^{2}}}
Thus, option (a) is the correct answer.

Note: Students should know the formula of differentiation and formulas related to it otherwise it will become difficult to solve and the answer will be incorrect. Also, remember we have taken log⁡e=log⁡ee=1\log e={{\log }_{e}}e=1 if we take this as log e base 10 then the answer will be some numeric value and the answer will be totally changed. So, be careful in assuming the things and in order to avoid mistakes.