What is the nth derivative of f(x) = x e^(2x)?

Apr 13, 2015

We can write the derivative of this function in a ricorsive way.
Ricorsive way means that also in the definition there is the concept of derivate!

${y}^{\left(1\right)} = 1 \cdot {e}^{2 x} + x \cdot {e}^{2 x} \cdot 2 = {e}^{2 x} + 2 x {e}^{2 x} = {e}^{2 x} + 2 y$

${y}^{\left(2\right)} = 2 {e}^{2 x} + 2 {y}^{\left(1\right)}$

${y}^{\left(3\right)} = {2}^{2} {e}^{2 x} + 2 {y}^{\left(2\right)}$

...

${y}^{\left(n\right)} = {2}^{n - 1} {e}^{2 x} + 2 {y}^{\left(n - 1\right)}$.

I hope it is the answer you wanted!

Apr 13, 2015

Thank you for providing a fun question to work on!

$f \left(x\right) = x {e}^{2 x}$
$f ' \left(x\right) = {e}^{2 x} + 2 x {e}^{2 x} = {e}^{2 x} \left(2 x + 1\right)$
$f ' ' \left(x\right) = 2 {e}^{2 x} \left(2 x + 1\right) + {e}^{2 x} \cdot 2 = 2 {e}^{2 x} \left(2 x + 2\right)$
$f ' ' ' \left(x\right) = 4 {e}^{2 x} \left(2 x + 2\right) + 2 {e}^{2 x} \cdot 2 = 4 {e}^{2 x} \left(2 x + 3\right)$

${f}^{\left(4\right)} \left(x\right) = 8 {e}^{2 x} \left(2 x + 3\right) + 8 {e}^{2 x} = 8 {e}^{2 x} \left(2 x + 4\right)$

Looks like

${f}^{\left(n\right)} \left(x\right) = {2}^{n - 1} {e}^{2 x} \left(2 x + n\right)$

It should be straightforward to prove by induction on $n$.