Question

If ${x^y} = {y^x}$, then $x(x - y\log x)\dfrac{{dy}}{{dx}}$ is equal to
\[1)\]$y(y - x\log y)$
\[2)\]$y(y + x\log y)$
\[3)\]$x(x + y\log x)$
\[4)\]$x(y - x\log y)$

Answer 1

Hint: We have to find the value of the given$x(x - y\log x)\dfrac{{dy}}{{dx}}$. And it’s given that ${x^y} = {y^x}$ . We solve the question using the given relation , the concept of logarithmic functions and the concept of differentiation of functions . We will solve this question using the chain rule and product rule .
The Product Rule : $\dfrac{{d(uv)}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$.

Complete step-by-step answer:
Differentiation , in mathematics , is the process of finding the derivative , or the rate of change of a given function . In contrast to the abstract nature of the theory behind it , the practical technique of differentiation can be carried out by purely algebraic manipulations , using three basic derivatives , four rules of operation , and a knowledge of how to manipulate functions . We can solve any of the problems using the rules of operations i.e. addition , subtraction , multiplication and division .
Given :
${x^y} = {y^x}$
Taking log both sides , we get
$\log \left( {{x^y}} \right) = \log \left( {{y^x}} \right)$----(1)
We know the basic formulas of log that ,
$\log \left( {{x^n}} \right) = n\log x$
So, the equation (1) can be expressed as:
$y\log x = x\log y$------(2)
Now , differentiating (2) with respect to ‘x’ , we get
Now , using the product rule mentioned above in the hint , we get
$\dfrac{{d}}{{dx}}\left( {y\log x} \right) = \dfrac{{d}}{{dx}}\left( {x\log y} \right)$
$\dfrac{{dy}}{{dx}}\log x + y \times \left( {\dfrac{1}{x}} \right) = \log y + x \times \left( {\dfrac{1}{y}} \right)\dfrac{{dy}}{{dx}}$
On simplifying , we get
$\dfrac{{dy}}{{dx}}\log x + \dfrac{y}{x} = \log y + \left( {\dfrac{x}{y}} \right)\dfrac{{dy}}{{dx}}$,
Taking the differential terms on same side of the equation
$\dfrac{{dy}}{{dx}}\log x - \left( {\dfrac{x}{y}} \right)\dfrac{{dy}}{{dx}} = \log y - \left( {\dfrac{y}{x}} \right)$
$\dfrac{{dy}}{{dx}}\left( {\log x - \dfrac{x}{y}} \right) = \log y - \left( {\dfrac{y}{x}} \right)$
Taking the LCM on both sides of the equation,
 $\dfrac{{dy}}{{dx}}\left( {\dfrac{{y\log x - x}}{y}} \right) = \left( {\dfrac{{x\log y - y}}{x}} \right) $
  $\dfrac{{dy}}{{dx}} = \left( {\dfrac{{\dfrac{{x\log y - y}}{x}}}{{\dfrac{{y\log x - x}}{y}}}} \right) \Rightarrow \left( {\dfrac{{x\log y - y}}{x}} \right) \times \left( {\dfrac{y}{{y\log x - x}}} \right) $
Taking (-1) common from both numerator and denominator , we get
$\dfrac{{dy}}{{dx}} = \left( {\dfrac{{y\left( {y - x\log y} \right)}}{{x\left( {x - y\log x} \right)}}} \right)$
$x\left( {x - y\log x} \right)\dfrac{{dy}}{{dx}} = y\left( {y - x\log y} \right)$
Thus , the value of $x\left( {x - y\log x} \right)\dfrac{{dy}}{{dx}}$ is $y\left( {y - x\log y} \right)$
Hence , the correct option is (1) .
So, the correct answer is “Option 1”.

Note: We differentiated y with respect to ‘x’ to find$\dfrac{{dy}}{{dx}}$.
We know the differentiation of various function :
\[\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x\]
\[\dfrac{{d\left[ {sin{\text{ }}x} \right]{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x\]
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = {sec^2}x$