Question

If \[{{y}^{x}}-{{x}^{y}}=1\] then the value of \[\dfrac{dy}{dx}\] at \[x=1\] is
(a) \[2\left( 1-\log 2 \right)\]
(b) \[2\left( 1+\log 2 \right)\]
(c) \[\left( 2-\log 2 \right)\]
(d) \[\left( 2+\log 2 \right)\]

Answer 1

Hint: We solve this problem by using the standard formulas of differentiation. When there are two functions \[u,v\] in the form \[{{u}^{v}}\] then the derivative is found by dividing the function \[{{u}^{v}}\]in two forms such that first considering \[u\] as constant and next \[v\] as some constant that is
\[\dfrac{d}{dx}\left( {{u}^{v}} \right)=\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering u as constant} \right)+\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering v as constant} \right)\]
Also, we use the formula of standard derivative as
\[\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}.\log a\]where \['a'\] is constant.

Complete step by step answer:
We are given that the equation as
\[\Rightarrow {{y}^{x}}-{{x}^{y}}=1\]
We are asked to find \[\dfrac{dy}{dx}\] at \[x=1\].
Let us find the value of \['y'\] for \[x=1\].
By substituting \[x=1\] in given equation we get
\[\begin{align}
  & \Rightarrow {{y}^{1}}-{{1}^{y}}=1 \\
 & \Rightarrow y=2 \\
\end{align}\]
Now, let us take the given equation as
\[\Rightarrow {{y}^{x}}-{{x}^{y}}=1\]
Now, by differentiating with respect to \['x'\] on both sides we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {{y}^{x}} \right)-\dfrac{d}{dx}\left( {{x}^{y}} \right)=\dfrac{d}{dx}\left( 1 \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( {{y}^{x}} \right)-\dfrac{d}{dx}\left( {{x}^{y}} \right)=0.........equation(i) \\
\end{align}\]
Let us evaluate the first derivative taking as
\[P=\dfrac{d}{dx}\left( {{y}^{x}} \right)\]
We know that when there are two functions \[u,v\] in the form \[{{u}^{v}}\] then the derivative is found by dividing the function \[{{u}^{v}}\]in two forms such that first considering \[u\] as constant and next \[v\] as some constant that is
\[\dfrac{d}{dx}\left( {{u}^{v}} \right)=\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering u as constant} \right)+\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering v as constant} \right)\]
By using the above formula to above equation we get
\[\begin{align}
  & \Rightarrow P=\dfrac{d}{dx}\left( {{y}^{x}}\text{ taing y as constant} \right)+\dfrac{d}{dx}\left( {{y}^{x}}\text{ taking x as constant} \right) \\
 & \Rightarrow P=x.{{y}^{x-1}}.\dfrac{dy}{dx}+\dfrac{d}{dx}\left( {{y}^{x}}\text{ taking y as constant} \right) \\
\end{align}\]
We know the standard formula that is
\[\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}.\log a\]where \['a'\] is constant.
By using this formula to above equation we get
\[\Rightarrow P=x.{{y}^{x-1}}.\dfrac{dy}{dx}+{{y}^{x}}.\log y\]
Now let us evaluate the second derivative from equation (ii) as
\[Q=\dfrac{d}{dx}\left( {{x}^{y}} \right)\]
By using the similar formulas from the above we get
\[\Rightarrow Q=y.{{x}^{y-1}}+{{x}^{y}}.\log x.\dfrac{dy}{dx}\]
Now, by substituting the values of \[P,Q\] in equation (i) we get
\[\Rightarrow \left( x.{{y}^{x-1}}.\dfrac{dy}{dx}+{{y}^{x}}.\log y \right)-\left( y.{{x}^{y-1}}+{{x}^{y}}.\log x.\dfrac{dy}{dx} \right)=0\]
Now, by rearranging the terms we get
\[\Rightarrow \dfrac{dy}{dx}\left( x.{{y}^{x-1}}-{{x}^{y}}.\log x \right)=y.{{x}^{y-1}}-{{y}^{x}}\log y\]
We are asked to find the value of \[\dfrac{dy}{dx}\] at \[x=1\].
Now, by substituting \[x=1,y=2\] in above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}\left( 1\times {{2}^{1-1}}-{{1}^{2}}\times \log 1 \right)=2\times {{1}^{2-1}}-{{2}^{1}}\log 2 \\
 & \Rightarrow \dfrac{dy}{dx}\left( 1-0 \right)=2-2\log 2 \\
 & \Rightarrow \dfrac{dy}{dx}=2\left( 1-\log 2 \right) \\
\end{align}\]
Therefore, we can say that\[\dfrac{dy}{dx}\] at \[x=1\] is \[2\left( 1-\log 2 \right)\].

So, the correct answer is “Option a”.

Note: Students may make mistakes in the derivative part only. We have the formula that is when there are two functions \[u,v\] in the form \[{{u}^{v}}\] then the derivative is found by dividing the function \[{{u}^{v}}\]in two forms such that first considering \[u\] as constant and next \[v\] as some constant that is
\[\dfrac{d}{dx}\left( {{u}^{v}} \right)=\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering u as constant} \right)+\dfrac{d}{dx}\left( {{u}^{v}}\text{ considering v as constant} \right)\]
The application of this formula gives confusion to students. Applying this formula with care leads to the correct answer. Only this part needs to be taken care of.