Question

What is the second derivative of $ f\left( x \right) = x{e^{{x^2}}} $ ?

Answer 1

Hint: In order to find the second derivative of the given function, first we need to find the first derivative. For that we can see that there are two functions, one is $ x $ and second is $ {e^{{x^2}}} $ , which is needed to be solved separately, so we would be using product rule to solve this, and our first derivative is obtained. Similarly, we obtain our first derivative result as a part of two functions, so we would again use product rule to solve, and our second derivative is obtained.

Complete step by step solution:
We are given with the function $ f\left( x \right) = x{e^{{x^2}}} $ , which is a set of two different functions that is $ x $ and $ {e^{{x^2}}} $ .
From product rule we know how to derive two functions separately in a multiplication, and the rule is as: $ \dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}} $ .
Comparing $ uv $ with our function $ f\left( x \right) = x{e^{{x^2}}} $ , we get that $ u = x $ and $ v = {e^{{x^2}}} $ .
Substituting these values in the product rule and we get:
 \[\dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = x\dfrac{{d{e^{{x^2}}}}}{{dx}} + {e^{{x^2}}}\dfrac{{dx}}{{dx}}\]
On further solving with formulas of derivatives, we get:
 \[
  \dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = x.\left( {2x{e^{{x^2}}}} \right) + {e^{{x^2}}} \\
  \dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = x.2x{e^{{x^2}}} + {e^{{x^2}}} \\
  \dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = 2{x^2}{e^{{x^2}}} + {e^{{x^2}}} \;
 \]
Taking \[{e^{{x^2}}}\] common:
 \[\dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = {e^{{x^2}}}\left( {2{x^2} + 1} \right)\]
Therefore, our first derivative obtained is: \[\dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}} = {e^{{x^2}}}\left( {2{x^2} + 1} \right)\]
Derivating the obtained first derivative again with respect to \[dx\] , to obtain our second derivative.
We can see that there are two functions present in first derivative also, so again using product rule:
Comparing $ uv $ with our function \[{e^{{x^2}}}\left( {2{x^2} + 1} \right)\] , we get that $ u = {e^{{x^2}}} $ and $ v = \left( {2{x^2} + 1} \right) $ .
Substituting these values in the product rule and we get:
 \[\dfrac{{d\left( {\dfrac{{d\left( {x{e^{{x^2}}}} \right)}}{{dx}}} \right)}}{{dx}} = {e^{{x^2}}}\dfrac{{d\left( {2{x^2} + 1} \right)}}{{dx}} + \left( {2{x^2} + 1} \right)\dfrac{{d{e^{{x^2}}}}}{{dx}}\]
On further solving with formulas of derivatives, we get:
 \[
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}\dfrac{{d\left( {2{x^2} + 1} \right)}}{{dx}} + \left( {2{x^2} + 1} \right)\dfrac{{d{e^{{x^2}}}}}{{dx}} \\
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}.\left( {\dfrac{{d2{x^2}}}{{dx}} + \dfrac{{d1}}{{dx}}} \right) + \left( {2{x^2} + 1} \right)\dfrac{{d{e^{{x^2}}}}}{{dx}} \\
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}.\left( {4x + 0} \right) + \left( {2{x^2} + 1} \right).2x{e^{{x^2}}} \;
 \]
Taking \[{e^{{x^2}}}\] common:
 \[
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}.\left( {4x + \left( {2{x^2} + 1} \right).2x} \right) \\
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}.\left( {4x + 4{x^3} + 2x} \right) \\
  \dfrac{{{d^2}\left( {x{e^{{x^2}}}} \right)}}{{d{x^2}}} = {e^{{x^2}}}.\left( {4{x^3} + 6x} \right) \;
 \]
The second derivative obtained is \[{e^{{x^2}}}.\left( {4{x^3} + 6x} \right)\] .
Therefore, the second derivative of $ f\left( x \right) = x{e^{{x^2}}} $ is \[{e^{{x^2}}}.\left( {4{x^3} + 6x} \right)\].
So, the correct answer is “ \[{e^{{x^2}}}.\left( {4{x^3} + 6x} \right)\]”.

Note: We could have also considered $f\left( x \right) = y$, for our first derivative it can be written as $f'\left( x \right)$ or $y'$. Similarly, for the second derivative, it's written as $f''\left( x \right)$ or $y''$.
It’s important to remember product rules to solve these types of questions.
The product rule: $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$