Question

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

Answer 1

Hint: Here, first we will convert the given equation ${{x}^{y}}={{e}^{x-y}}$ in form of log by using the formula $n\log m=\log {{m}^{n}}$ . Then, we will have an equation like this $y\log x=\left( x-y \right)$ . We will then differentiate this with respect to x. We will be using the formula $\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+\dfrac{dx}{dx}\cdot y$ , $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ , $\dfrac{d}{dx}x=1$ . Thus, on solving we will get the answer.

Complete step-by-step solution:
Here, we are given the equation ${{x}^{y}}={{e}^{x-y}}$ . So, we will use the formula $n\log m=\log {{m}^{n}}$ .
We will first multiply with $\log $ on both the sides. So, we get as
$\log {{x}^{y}}=\log {{e}^{x-y}}$
Now, we will apply the formula $n\log m=\log {{m}^{n}}$ so, we get as
$y\log x=\left( x-y \right)\log e$
Now, we will assume that $\log e={{\log }_{e}}e=1$ because we know the formula ${{\log }_{n}}n=1$ . So, we get equation as
$y\log x=\left( x-y \right)$ ……………..(1)
Now, we will differentiate the equation with respect to x. We can write it as
$\dfrac{d}{dx}\left( y\log x \right)=\dfrac{d}{dx}\left( x-y \right)$
Here, we will use the product rule i.e. given as $\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+\dfrac{dx}{dx}\cdot y$ . So, here we will take x as y and y as $\log x$ . So, we can write it as
$y\dfrac{d}{dx}\log x+\dfrac{dy}{dx}\cdot \log x=\dfrac{d}{dx}x-\dfrac{dy}{dx}$
Now, we know that differentiation of $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ , $\dfrac{d}{dx}x=1$ . So, we can write it as
$y\cdot \dfrac{1}{x}+\dfrac{dy}{dx}\cdot \log x=1-\dfrac{dy}{dx}$
Now, taking $\dfrac{dy}{dx}$ terms on left side and other terms on right side, we get as
$\dfrac{dy}{dx}+\dfrac{dy}{dx}\cdot \log x=1-y\cdot \dfrac{1}{x}$
On taking $\dfrac{dy}{dx}$ common, we get as
$\dfrac{dy}{dx}\left( 1+\log x \right)=1-\dfrac{y}{x}$
On further simplification, we can write it as
$\dfrac{dy}{dx}=\dfrac{x-y}{x\left( 1+\log x \right)}$
Now, from equation (1), we will put value of $x-y$ so, we get as
$\dfrac{dy}{dx}=\dfrac{y\log x}{x\left( 1+\log x \right)}$ ………………..(2)
Now, from equation (1) i.e. $y\log x=\left( x-y \right)$ we can write it as $y\log x+y=x$ . On taking y common we get as
$y\left( \log x+1 \right)=x$
$\dfrac{y}{x}=\dfrac{1}{\left( \log x+1 \right)}$
We will substitute this value in the equation (2) so, we get as
$\dfrac{dy}{dx}=\dfrac{\log x}{{{\left( 1+\log x \right)}^{2}}}$
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 $\log e={{\log }_{e}}e=1$ 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.