Question

If we have a function as $y=x\log \left( \dfrac{x}{a+bx} \right)$ , then ${{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=$
1. $x\dfrac{dy}{dx}-y$
2. ${{\left( x\dfrac{dy}{dx}-y \right)}^{2}}$
3. $y\dfrac{dy}{dx}-x$
4. ${{\left( y\dfrac{dy}{dx}-x \right)}^{2}}$

Answer 1

Hint: For solving this question you should know about solving the derivative of a composite function because it is a composite function. And this is a logarithmic function also. So, for solving this we will divide it in two parts and then differentiate and then at the end substitute the values in the chain rule. And thus we will get our answer.

Complete step-by-step solution:
According to the question we have to find the value of ${{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ and the function is given to us as $y=x\log \left( \dfrac{x}{a+bx} \right)$. As we know that the first derivative is $\dfrac{dy}{dx}$ and the second derivative will be $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. Here it is clear that it,
$\begin{align}
  & \Rightarrow y=x\log \left( \dfrac{x}{a+bx} \right) \\
 & \Rightarrow \dfrac{y}{x}=\log x-\log \left( a+bx \right) \\
\end{align}$
Differentiating with respect to $x$, we will get as follows,
$\dfrac{x.\dfrac{dy}{dx}-y}{{{x}^{2}}}=\dfrac{1}{x}-\dfrac{b}{a+bx}$
Or,
$x.\dfrac{dy}{dx}-y=x-\dfrac{b{{x}^{2}}}{a+bx}$
Differentiating again with respect to $x$, we will get as follows,
\[\begin{align}
  & x.\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}-\dfrac{dy}{dx}=1-\dfrac{\left[ \left( a+bx \right)2bx-b{{x}^{2}}\left( b \right) \right]}{{{\left( a+bx \right)}^{2}}} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\left( a+bx \right)}^{2}}-2bx\left( a+bx \right)+{{b}^{2}}{{x}^{2}}}{{{\left( a+bx \right)}^{2}}} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( a+bx \right)\left[ a-bx-2bx \right]+{{b}^{2}}{{x}^{2}}}{{{\left( a+bx \right)}^{2}}} \\
\end{align}\]
On further calculations, we will get,
\[\begin{align}
  & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}-{{b}^{2}}{{x}^{2}}+{{b}^{2}}{{x}^{2}}}{{{\left( a+bx \right)}^{2}}} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}}{{{\left( a+bx \right)}^{2}}} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( \dfrac{a}{a+bx} \right)}^{2}}\ldots \ldots \ldots \left( i \right) \\
 & \\
\end{align}\]
Differentiating the given expression with respect to $x$, we will get,
$x\dfrac{dy}{dx}=x\dfrac{\left( a+bx-bx \right)}{{{\left( a+bx \right)}^{2}}}\left( \dfrac{a+bx}{x} \right)+\log \left( \dfrac{x}{a+bx} \right)$
Or, we can say,
$\dfrac{dy}{dx}=\dfrac{a}{a+bx}+\dfrac{y}{x}$
Or as follows,
$\dfrac{dy}{dx}-\dfrac{y}{x}=\dfrac{a}{a+bx}\ldots \ldots \ldots \left( ii \right)$
From equations (i) and (ii), we will get as follows,
$\begin{align}
  & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( \dfrac{dy}{dx}-\dfrac{y}{x} \right)}^{2}} \\
 & \Rightarrow x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}}{{\left( x\dfrac{dy}{dx}-y \right)}^{2}} \\
 & \Rightarrow {{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( x\dfrac{dy}{dx}-y \right)}^{2}} \\
\end{align}$
So, the correct answer to the given expression is option 2.

Note: If you find the derivative of any composite function then you should always apply the chain rule there because it makes our calculations easy and accurate. And if we don’t apply this then we will have to differentiate the composite function very carefully and each and every term should be differentiated.