Question

For \[(ax + b){e^{\left( {\dfrac{y}{x}} \right)}} = x\] . Find $\dfrac{{{d^2}y}}{{d{x^2}}}$

Answer 1

Hint: In this question, we need to evaluate the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$. For this, we will apply logarithm on both sides of the equation. Express y in terms of x. Then find $\dfrac{{dy}}{{dx}}$. Take the derivative of $\dfrac{{dy}}{{dx}}$ to get the answer.

Complete step by step solution:
We are given that \[(ax + b){e^{\left( {\dfrac{y}{x}} \right)}} = x\].
We need to find the second derivative.
This equation can also be written as \[{e^{\left( {\dfrac{y}{x}} \right)}} = \dfrac{x}{{(ax + b)}}\]
To simplify this equation, we will apply logarithm on both sides.
Thus, we get
 \[\log {e^{\left( {\dfrac{y}{x}} \right)}} = \log (\dfrac{x}{{ax + b}})\]
We know that $\log {b^a} = a\log b$.
Therefore,
  \[\dfrac{y}{x}\log e = \log (\dfrac{x}{{ax + b}})\]
Now, $\log e = 1$.
Thus, we get
  \[
  \dfrac{y}{x} = \log (\dfrac{x}{{ax + b}}) \\
   \Rightarrow y = x\log (\dfrac{x}{{ax + b}})......(1) \;
 \]
Differentiating both sides of (1) with respect to x, we get
 \[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x\log (\dfrac{x}{{ax + b}}))\]
The product rule of derivatives is as follows:
If $y = u \times v$, then $\dfrac{{dy}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$.
Applying the product rule, we get
 \[\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}(\log (\dfrac{x}{{ax + b}})) + \log (\dfrac{x}{{ax + b}})\dfrac{d}{{dx}}(x)\]
We know that \[\dfrac{d}{{dx}}(x) = 1\] .
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}(\log (\dfrac{x}{{ax + b}})) + \log (\dfrac{x}{{ax + b}})......(2)\]
Also, $\log \dfrac{a}{b} = \log a - \log b$.
Therefore, \[\log (\dfrac{x}{{ax + b}}) = \log x - \log (ax + b)\] .
This implies that
  \[
  \dfrac{d}{{dx}}(\log (\dfrac{x}{{ax + b}})) \\
   = \dfrac{d}{{dx}}(\log x - \log (ax + b)) \\
   = \dfrac{d}{{dx}}(\log x) - \dfrac{d}{{dx}}(\log (ax + b)) \;
 \]
We know that \[\log (ax + b) = \dfrac{a}{{ax + b}}\] and $\log x = \dfrac{1}{x}$.
Therefore, we get \[\dfrac{d}{{dx}}(\log (\dfrac{x}{{ax + b}})) = \dfrac{1}{x} - \dfrac{a}{{ax + b}}.....(3)\] .
On combining, (2) and (3), we have
 \[\dfrac{{dy}}{{dx}} = x(\dfrac{1}{x} - \dfrac{a}{{ax + b}}) + \log (\dfrac{x}{{ax + b}})\]
Multiplying $x$on both the sides, we get
 \[x\dfrac{{dy}}{{dx}} = {x^2}(\dfrac{1}{x} - \dfrac{a}{{ax + b}}) + x\log (\dfrac{x}{{ax + b}})\]
Using (1), we have
  \[
  x\dfrac{{dy}}{{dx}} = {x^2}(\dfrac{1}{x} - \dfrac{a}{{ax + b}}) + y \\
   \Rightarrow x\dfrac{{dy}}{{dx}} = {x^2}(\dfrac{b}{{x(ax + b)}}) + y = \dfrac{{bx}}{{(ax + b)}} + y......(4) \;
 \]
We will again differentiate (4) with respect to x.
 \[
  x\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(\dfrac{{bx}}{{(ax + b)}}) + \dfrac{{dy}}{{dx}} \\
   \Rightarrow x\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(\dfrac{{bx}}{{(ax + b)}}) \;
 \]
The quotient rule of derivatives is as follows:
If $y = \dfrac{u}{v}$, then $\dfrac{{dy}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$.
On applying the quotient rule, we get
 \[
  x\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}(\dfrac{{bx}}{{(ax + b)}}) \\
   = \dfrac{{(ax + b)\dfrac{{d(bx)}}{{dx}} - bx\dfrac{{d(ax + b)}}{{dx}}}}{{{{(ax + b)}^2}}} \\
   = \dfrac{{b(ax + b) - abx}}{{{{(ax + b)}^2}}} \\
   = \dfrac{{abx + {b^2} - abx}}{{{{(ax + b)}^2}}} \\
   = \dfrac{{{b^2}}}{{{{(ax + b)}^2}}} \;
 \]
Thus, \[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{x} \times \dfrac{{{b^2}}}{{{{(ax + b)}^2}}} = \dfrac{{{b^2}}}{{x{{(ax + b)}^2}}}\]
Hence, the required answer is \[\dfrac{{{b^2}}}{{x{{(ax + b)}^2}}}\] .
So, the correct answer is “$\dfrac{{{b^2}}}{{x{{(ax + b)}^2}}}$”.

Note: In any equation related to derivatives which contains $e$, the first approach should be applying logarithm to eliminate $e$. This will simplify the equation. Only then proceed with the differentiation part like we did in the above question.