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

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

& \Rightarrow {{y}^{1}}-{{1}^{y}}=1 \\\ & \Rightarrow y=2 \\\ \end{aligned}$$ Now, let us take the given equation as $$\Rightarrow {{y}^{x}}-{{x}^{y}}=1$$ Now, by differentiating with respect to $$'x'$$ on both sides we get $$\begin{aligned} & \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{aligned}$$ Let us evaluate the first derivative taking as $$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{aligned} & \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{aligned}$$ We know the standard formula that is $$\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{aligned} & \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{aligned}$$ Therefore, we can say that$$\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.