Question

The gradient of the curve ${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}$ is given by
(a) $\dfrac{x}{y}$
(b) $\dfrac{y}{x}$
(c) $\dfrac{-x}{y}$
(d) $\dfrac{-y}{x}$

Answer 1

Hint: In this question, we are asked to find the gradient of a curve whose equation is given. As in two dimensions, the gradient of the curve should be equal to $\dfrac{dy}{dx}$, therefore, we should try to simplify the equation by taking logarithms on both sides of the equation and then use the chain rule to obtain a differential equation. Then, we can take terms involving $\dfrac{dy}{dx}$ to one side and simplify the resulting equation to find the value of $\dfrac{dy}{dx}$ which will be our required answer.

Complete step by step solution:
The given equation of the curve is
${{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}.................(1.1)$
As there are exponents in the expression, we can simplify the equation by taking logarithm on both sides. We can use the following properties of logarithms for any two numbers a and b
$\begin{align}
  & \log \left( ab \right)=\log \left( a \right)+\log \left( b \right) \\
 & \log \left( {{a}^{b}} \right)=b\log (a)......................(1.2) \\
\end{align}$
Using the properties in (1.2) and taking log on both sides of (1.1), we obtain
$\begin{align}
  & {{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}} \\
 & \Rightarrow \log \left( {{x}^{m}}{{y}^{n}} \right)=\log \left( {{\left( x+y \right)}^{m+n}} \right) \\
 & \Rightarrow m\log \left( x \right)+n\log \left( y \right)=(m+n)\log \left( x+y \right)..........(1.3) \\
\end{align}$
Now, we can use the chain rule of derivatives which states that if y is a function of x and f(y) is a function of y, then
$\dfrac{df(y)}{dx}=\dfrac{df(y)}{dy}\times \dfrac{dy}{dx}.....................(1.4)$
And the derivative of log function is given by
$\dfrac{d\log (x)}{dx}=\dfrac{1}{x}................(1.5)$
Taking derivative with respect to x on both sides of (1.3) and using equations (1.4) and (1.5), we get
$\begin{align}
  & m\log \left( x \right)+n\log \left( y \right)=(m+n)\log \left( x+y \right) \\
 & \Rightarrow \dfrac{d\left( m\log \left( x \right)+n\log \left( y \right) \right)}{dx}=\dfrac{d\left( (m+n)\log \left( x+y \right) \right)}{dx} \\
 & \Rightarrow m\dfrac{d\log x}{dx}+n\dfrac{d\log y}{dy}\times \dfrac{dy}{dx}=(m+n)\dfrac{d\log (x+y)}{d(x+y)}\times \dfrac{d(x+y)}{dx} \\
 & \Rightarrow \dfrac{m}{x}+\dfrac{n}{y}\times \dfrac{dy}{dx}=(m+n)\dfrac{1}{\left( x+y \right)}\left( \dfrac{dx}{dx}+\dfrac{dy}{dx} \right)=(m+n)\dfrac{1}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right) \\
\end{align}$
We can collect terms involving $\dfrac{dy}{dx}$ to the left hand side in the last line to obtain
$\begin{align}
  & \dfrac{dy}{dx}\times \left( \dfrac{n}{y}-\dfrac{m+n}{\left( x+y \right)} \right)=(m+n)\dfrac{1}{\left( x+y \right)}-\dfrac{m}{x} \\
 & \Rightarrow \dfrac{dy}{dx}\times \left( \dfrac{nx+ny-my-ny}{y\left( x+y \right)} \right)=\dfrac{mx+nx-mx-my}{x\left( x+y \right)} \\
 & \Rightarrow \dfrac{dy}{dx}\times \left( \dfrac{nx-my}{y\left( x+y \right)} \right)=\dfrac{nx-my}{x\left( x+y \right)} \\
\end{align}$
Cancelling the term $\dfrac{nx-my}{(x+y)}$ from both sides, we obtain
$\dfrac{dy}{dx}=\dfrac{y}{x}$
Which matches option (b) of the question. Therefore, option (b) is the correct answer to this question.

Note: We should note that we took logarithm on both sides to make the equations simpler to handle. However, we could have directly taken derivative with respect to x on both sides of the given equation and used the formula $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$ and the chain rule to find the value of $\dfrac{dy}{dx}$. However, the final answer will remain the same in both the methods.