Question

If $y = {x^2}{e^{mx}}$ where, $m$ is a constant then, $\dfrac{{{d^3}y}}{{d{x^3}}}$ is equal to
A.$m{e^{mx}}\left( {6mx + 6 + {m^2}{x^2}} \right)$
B.$2{m^3}x{e^{mx}}$
C.$m{e^{mx}}\left( {2mx + 2 + {m^2}{x^2}} \right)$
D.None of these

Answer 1

Hint: The given question contains an expression. We are supposed to find out the third order derivative of the equation. That is, we need to differentiate the given expression three times. We will differentiate the given expression once. Then we differentiate the answer and so on.
The formulas used to solve the problem are:
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ and $\dfrac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x}$

Complete answer: The given expression is,
$y = {x^2}{e^{mx}}$
Now, we know the product rule of differentiation $\dfrac{{d\left[ {f\left( x \right) \times g\left( x \right)} \right]}}{{dx}} = f\left( x \right)g'\left( x \right) + f'\left( x \right)g\left( x \right)$. So, we get,
Now, we differentiate with respect to $x$, using the formulas mentioned above
$y' = \left( {{e^{mx}}} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right) + \left( {{x^2}} \right)\dfrac{d}{{dx}}\left( {{e^{mx}}} \right)$
$ \Rightarrow y' = {e^{mx}} \times 2x + {x^2}\left( {m{e^{mx}}} \right) = 2x{e^{mx}} + m{x^2}{e^{mx}}$
Taking out ${e^{mx}}$common,
$ \Rightarrow y' = {e^{mx}}\left( {2x + m{x^2}} \right)$
This is the first order derivative.
Now, we need to find out the second order derivative.
That is to differentiate $y'$ with respect to $x$. So, we get,
$ \Rightarrow y'' = {e^{mx}}\dfrac{d}{{dx}}\left( {2x + m{x^2}} \right) + \left( {2x + m{x^2}} \right)\dfrac{d}{{dx}}\left( {{e^{mx}}} \right)$
$ \Rightarrow y'' = {e^{mx}}\left( {2 + 2mx} \right) + \left( {2x + m{x^2}} \right)\left( {m{e^{mx}}} \right)$
${e^{mx}}$ is a common term, therefore, we get,
\[ \Rightarrow y'' = {e^{mx}}\left[ {2 + 2mx + 2mx + {m^2}{x^2}} \right]\]
Now, by adding the similar terms and rearranging the equation, we get,
\[ \Rightarrow y'' = {e^{mx}}\left[ {2 + 4mx + {m^2}{x^2}} \right]\]
This is the second order derivative.
Next, we move on to the last step that is finding the third order derivative.
We differentiate $y''$ with respect to $x$
$ \Rightarrow y''' = {e^{mx}}\dfrac{d}{{dx}}\left[ {2 + 4mx + {m^2}{x^2}} \right] + \left[ {2 + 4mx + {m^2}{x^2}} \right]\dfrac{d}{{dx}}\left( {{e^{mx}}} \right)$
Taking $m$ common from all the terms, we get,
$ \Rightarrow y''' = {e^{mx}}\left[ {4m + 2{m^2}x} \right] + \left[ {2 + 4mx + {m^2}{x^2}} \right]\left( {m{e^{mx}}} \right)$
Grouping and rearranging the equations,
$ \Rightarrow y''' = m{e^{mx}}\left[ {4 + 2mx + 2 + 4mx + {m^2}{x^2}} \right]$
Adding the similar terms,
$ \Rightarrow y''' = m{e^{mx}}\left[ {6 + 6mx + {m^2}{x^2}} \right]$
Therefore, the final answer is $y''' = m{e^{mx}}\left[ {6 + 6mx + {m^2}{x^2}} \right]$
Hence the correct answer is option A.

Note:
The question does not mention in words that we need to differentiate thrice, but we need to understand that $\dfrac{{{d^3}y}}{{d{x^3}}}$ means differentiating three times, since the order of the derivative is 3. While differentiating the second and the third time, be very mindful because the terms are more. So, solve it step by step so that there is no confusion.