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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.
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.
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