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Question: If \[{x^y} = {e^{x - y}}\], then \[\dfrac{{dy}}{{dx}}\] is equal to (A) \[\dfrac{{\log x}}{{1 + \l...

If xy=ex−y{x^y} = {e^{x - y}}, then dydx\dfrac{{dy}}{{dx}} is equal to
(A) log⁡x1+log⁡x\dfrac{{\log x}}{{1 + \log x}}
(B) log⁡x1−log⁡x\dfrac{{\log x}}{{1 - \log x}}
(C) log⁡x(1+log⁡x)2\dfrac{{\log x}}{{{{\left( {1 + \log x} \right)}^2}}}
(D) ylog⁡xx(1+log⁡x)2\dfrac{{y\log x}}{{x{{\left( {1 + \log x} \right)}^2}}}

Explanation

Solution

In this question, we have to choose the required solution from the given particular options. The given of the question is the value of the given term, it is an exponential function. To find the required solution, first we have to logarithm both sides and then we will differentiate both sides with respect to x using product rule of differentiation. After that some calculations we will get the required result.

Formula used: Product Rule:
If u and v are two functions of x, then the derivative of the product uvuv is given by: d(uv)dx=udvdx+vdudx\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}.
Logarithm formula:
log⁡(bx)=x.log⁡(b)\log \left( {{b^x}} \right) = x.\log \left( b \right)
log⁡(ex)=x\log \left( {{e^x}} \right) = x

Complete step-by-step answer:
It is given that, xy=ex−y{x^y} = {e^{x - y}}.
We need to find out the value of dydx\dfrac{{dy}}{{dx}}.
Now, xy=ex−y{x^y} = {e^{x - y}}
Taking logarithm on both sides we get,
\Rightarrow$$$\log \left( {{x^y}} \right) = \log \left( {{e^{x - y}}} \right)$$ \Rightarrowy\log \left( x \right) = x - y$$…………....i) Differentiating both sides with respect to x, we get, $\Rightarrow\dfrac{{dy}}{{dx}}\log x + y\dfrac{1}{x} = 1 - \dfrac{{dy}}{{dx}}Taking Taking\dfrac{{dy}}{{dx}} in the same side, we get, $\Rightarrow$$$\dfrac{{dy}}{{dx}}\log x + \dfrac{{dy}}{{dx}} = 1 - y\dfrac{1}{x}
Taking common terms outside from the brackets,
\Rightarrow$$$\left( {\log x + 1} \right)\dfrac{{dy}}{{dx}} = 1 - \dfrac{y}{x}$$ Simplifying we get, \Rightarrow\dfrac{{dy}}{{dx}} = \dfrac{{1 - \dfrac{y}{x}}}{{\left( {\log x + 1} \right)}}$$………...…….ii) From i) we can write, $$y\log \left( x \right) + y = x$$ $\Rightarrowy\left( {1 + \log x} \right) = xTherefore, Therefore,\dfrac{y}{x} = \dfrac{1}{{\left( {1 + \log x} \right)}} Hence we get from ii) $\Rightarrow$$$\dfrac{{dy}}{{dx}} = \dfrac{{1 - \dfrac{1}{{1 + \log x}}}}{{\left( {\log x + 1} \right)}} = \dfrac{{\dfrac{{1 + \log x - 1}}{{1 + \log x}}}}{{1 + \log x}} = \dfrac{{\log x}}{{{{\left( {1 + \log x} \right)}^2}}}
Thus we get,
⇒\Rightarrowdydx=log⁡x(1+log⁡x)2\dfrac{{dy}}{{dx}} = \dfrac{{\log x}}{{{{\left( {1 + \log x} \right)}^2}}}

∴\therefore The option C is the correct option.

Note: The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.