MCQ
If $y = {x^2}{e^{mx}}$, where $m$ is a constant, then ${{{d^3}y} \over {d{x^3}}} = $
- ✓$m{e^{mx}}({m^2}{x^2} + 6mx + 6)$
- B$2{m^3}x{e^{mx}}$
- C$m{e^{mx}}({m^2}{x^2} + 2mx + 2)$
- DNone of these
Differentiating w.r.t. $x$, we get
$\frac{{dy}}{{dx}} = 2x{e^{mx}} + m{x^2}{e^{mx}}$
Again, $\frac{{{d^2}y}}{{d{x^2}}} = 2({e^{mx}} + mx{e^{mx}}) + m(2x{e^{mx}} + {x^2}m{e^{mx}})$
or $\frac{{{d^2}y}}{{d{x^2}}} = {e^{mx}}({m^2}{x^2} + 4mx + 2)$
Again, $\frac{{{d^3}y}}{{d{x^3}}} = {e^{mx}}[{m^3}{x^2} + 4{m^2}x + 2m + 2{m^2}x + 4m]$
$ = {e^{mx}}[{m^3}{x^2} + 6{m^2}x + 6m]$
$ \Rightarrow \frac{{{d^3}y}}{{d{x^3}}} = m{e^{mx}}({m^2}{x^2} + 6mx + 6)$.
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