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.