Question

If x and y are related as ${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$, then prove that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$.

Answer 1

Hint: In this question, we are given the relation between x and y in terms of exponents, therefore, we should first try to take logarithms on both sides to convert the equation into a linear equation. Thereafter, we can use the properties of logarithm and the formula for derivative of logarithm to obtain the required answer.

Complete step-by-step solution -
The given relation is ${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$. Now, we know that for two numbers a and b, the properties of logarithmic functions are
$\log \left( {{a}^{b}} \right)=b\log \left( a \right)................(1.1)$
And $\log \left( ab \right)=\log \left( a \right)+\log \left( b \right)................(1.2)$
Also, the chain rule of derivatives states that for a function f depending on another function g of x
$\dfrac{df}{dx}=\dfrac{df}{dg}\times \dfrac{dg}{dx}......................(1.3)$
Now, we can take logarithms on both sides of the given relation and then use equation (1.1) and equation (1.2) to obtain
$\begin{align}
  & \log \left( {{x}^{m}}{{y}^{n}} \right)=\log \left( {{\left( x+y \right)}^{m+n}} \right) \\
 & \Rightarrow \log ({{x}^{m}})+\log ({{y}^{n}})=\left( m+n \right)\log \left( x+y \right) \\
 & \Rightarrow m\log x+n\log y=\left( m+n \right)\log \left( x+y \right) \\
\end{align}$
Now, we can take the derivative of both sides with respect to x and use the chain rule as in equation (1.3) with f as the logarithmic function and g as the function inside the parenthesis to obtain
$\begin{align}
  & \dfrac{d\left( m\log x+n\log y \right)}{dx}=\dfrac{d\left( \left( m+n \right)\log \left( x+y \right) \right)}{dx} \\
 & \Rightarrow m\dfrac{d\left( \log x \right)}{dx}+n\dfrac{d\left( \log y \right)}{dx}=(m+n)\dfrac{d\left( \log \left( x+y \right) \right)}{dx}..............(1.4) \\
\end{align}$
Now, we can use the formula for derivatives of log function as
$\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}.................(1.5)$
And $\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( \log y \right)}{dy}\times \dfrac{dy}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}..............(1.6)$
And then the chain rule to express $\begin{align}
  & \dfrac{d\left( \log \left( x+y \right) \right)}{dx}=\dfrac{d\left( \log \left( x+y \right) \right)}{d\left( x+y \right)}\dfrac{d\left( x+y \right)}{dx}=\dfrac{d\left( \log \left( x+y \right) \right)}{d\left( x+y \right)}\times \left( \dfrac{d\left( x \right)}{dx}+\dfrac{d\left( y \right)}{dx} \right) \\
 & \Rightarrow \dfrac{d\left( \log \left( x+y \right) \right)}{dx}=\dfrac{1}{\left( x+y \right)}\times \left( 1+\dfrac{dy}{dx} \right)................(1.7) \\
\end{align}$
Thus, using (1.5), (1.6) and (1.7) in equation (1.4), we get
$\begin{align}
  & m\dfrac{d\left( \log x \right)}{dx}+n\dfrac{d\left( \log y \right)}{dx}=(m+n)\dfrac{d\left( \log \left( x+y \right) \right)}{dx} \\
 & \Rightarrow m\times \dfrac{1}{x}+n\times \dfrac{1}{y}\times \dfrac{dy}{dx}=\left( m+n \right)\dfrac{1}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right) \\
\end{align}$
Collecting $\dfrac{dy}{dx}$ into the left hand side and taking other terms into the Right Hand Side(RHS), we obtain
$\begin{align}
  & \dfrac{dy}{dx}\times \left( \dfrac{n}{y}-\dfrac{m+n}{x+y} \right)=\dfrac{m+n}{\left( x+y \right)}-\dfrac{m}{x} \\
 & \Rightarrow \dfrac{dy}{dx}\times \left( \dfrac{n(x+y)-y(m+n)}{y(x+y)} \right)=\dfrac{x(m+n)-m(x+y)}{x(x+y)} \\
 & \Rightarrow \dfrac{dy}{dx}\times \left( \dfrac{nx-ym}{y(x+y)} \right)=\dfrac{xn-ym}{x(x+y)} \\
 & \\
\end{align}$
Therefore, we can now cancel out $xn-ym$ from the numerator of both sides and $(x+y)$ from the denominator of both sides to get
$\begin{align}
  & \dfrac{dy}{dx}\times \dfrac{1}{y}=\dfrac{1}{x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}...................(1.8) \\
\end{align}$
We also know that the product and quotient rule of differentiation is given by
\[\begin{align}
  & \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.....................(1.8a) \\
 & \dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}.................(1.8b) \\
\end{align}\]
Taking u=1 in (1.8b), we obtain
$\dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{x\times \dfrac{d\left( 1 \right)}{dx}-1\times \dfrac{d\left( x \right)}{dx}}{{{x}^{2}}}=\dfrac{-1}{{{x}^{2}}}.................(1.8c)$
Now, as we have to find the second order derivative, we can again differentiate (1.8) and use the quotient rule in (1.8b) to obtain
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{y}{x} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( \dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( \dfrac{y}{x} \right)}{dx}=\dfrac{x\dfrac{dy}{dx}-y\dfrac{dx}{dx}}{{{x}^{2}}}=\dfrac{x\times \dfrac{y}{x}-y\times 1}{{{x}^{2}}}=\dfrac{y-y}{{{x}^{2}}}=0...................(1.9) \\
\end{align}$
Where in the third term of the above equation, we have used $\dfrac{dy}{dx}=\dfrac{y}{x}$ and we have thus proved that
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$ which was asked in the question.

Note: We should note that we could also have found out the derivatives without taking the logarithm in both sides as in equation (1.4), but using the formulas for the derivatives of the power of functions. However, the answer obtained in both the methods will be the same and the calculations would be more complicated than was done by using logarithms.