Question

If ${X^Y} + {Y^X} = {a^b}$ , then find $\dfrac{{dy}}{{dx}}$ .

Answer 1

Hint: In this question we have to find out $\dfrac{{dy}}{{dx}}$ so, firstly we have to change these exponential terms into easier form so that it can be easily differentiated. So, use the log function to convert it into an easy format and differentiate with the following chain rule of differentiation.

Complete step-by-step solution:
Given: ${X^Y} + {Y^X} = {a^b}$ , then find $\dfrac{{dy}}{{dx}}$ .
In this question differentiation rules will be followed, so when we differentiate some terms with respect to $dx$ then chain rule will be followed.
So, $\because {X^Y} + {Y^X} = {a^b}$
Now, differentiate both sides with the help of $dx$
$\therefore \dfrac{{d\left( {{X^Y} + {Y^X}} \right)}}{{dx}} = \dfrac{{d{a^b}}}{{dx}}$
$ \Rightarrow \dfrac{d}{{dx}}({X^Y}) + \dfrac{d}{{dx}}({Y^X}) = 0$
Let ${X^Y} = A$ and ${Y^X} = B$
Applying both sides log function:
$\therefore Y\log X = \log A$ and $X\log Y = \log B.$
Now, differentiate both sides with respect to $dx$ , we have
$
   \Rightarrow Y\dfrac{{d\log X}}{{dx}} + \log X\dfrac{{dy}}{{dx}} = \dfrac{{d\log A}}{{dx}} \times \dfrac{{dA}}{{dx}} \\
   \Rightarrow \dfrac{Y}{X} + \operatorname{l} ogX\dfrac{{dy}}{{dx}} = \dfrac{1}{A} \times \dfrac{{dA}}{{dx}} \\
   \Rightarrow \dfrac{{dA}}{{dx}} = A\left( {\dfrac{Y}{X} + \log X\dfrac{{dy}}{{dx}}} \right) \\
  \therefore \dfrac{{dA}}{{dx}} = A\left( {\dfrac{Y}{X} + \log X\dfrac{{dy}}{{dx}}} \right) \to (1) \\
 $
Now, $\because \log B = X\log Y$
$
   \Rightarrow \dfrac{{d\log B}}{{dB}} \times \dfrac{{dB}}{{dx}} = X\dfrac{{d\log Y}}{{dy}} \times \dfrac{{dy}}{{dx}} + \log Y\dfrac{{dx}}{{dx}} \\
   \Rightarrow \dfrac{1}{B} \times \dfrac{{dB}}{{dx}} = \dfrac{x}{y} \times \dfrac{{dy}}{{dx}} + \log Y \\
  \therefore \dfrac{{dB}}{{dx}} = B\left( {\dfrac{x}{y} \times \dfrac{{dy}}{{dx}} + \log Y} \right) \to (2) \\
 $
Now, adding equation (1) and equation (2)
$
  \because \dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}} \Rightarrow \dfrac{d}{{dx}}({X^Y}) + \dfrac{d}{{dx}}({Y^X}) = 0 \\
   \Rightarrow \dfrac{{d{X^Y}}}{{dx}} + \dfrac{d}{{dx}}({Y^X}) \Rightarrow A\left( {\dfrac{y}{x} + \log X\dfrac{{dy}}{{dx}}} \right) + B\left( {\dfrac{x}{y} \times \dfrac{{dy}}{{dx}} + \log Y} \right) = 0 \\
 $
Now, putting the value of $A$ and $B$
$
  \because {X^Y}\left( {\dfrac{y}{x} + \log X\dfrac{{dy}}{{dx}}} \right) + {Y^x}\left( {\dfrac{x}{y} \times \dfrac{{dy}}{{dx}} + \log Y} \right) = 0 \\
   \Rightarrow Y{X^{Y - 1}} + \dfrac{{dy}}{{dx}}\left( {{X^Y}\log X + X{Y^{X - 1}}} \right) + {Y^X}\log Y = 0 \\ $

 $ \therefore \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {{Y^X}\log Y + Y{X^{Y - 1}}} \right)}}{{\left( {{X^Y}\log X + X{Y^{X - 1}}} \right)}} $
Hence, the value of $\dfrac{{dy}}{{dx}}$
$\dfrac{{dy}}{{dx}} = - \dfrac{{\left( {{Y^X}\log Y + Y{X^{Y - 1}}} \right)}}{{\left( {{X^Y}\log X + X{Y^{X - 1}}} \right)}}$

Note: In the question exponential terms must be changed into logarithmic terms. It will help to find out the value of $\dfrac{{dy}}{{dx}}$ easily and one more thing students should always use proper differentiation with the help of chain rule otherwise the answer will be wrong.