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

2 Answers
Massimiliano
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^((1))=1*e^(2x)+x*e^(2x)*2=e^(2x)+2xe^(2x)=e^(2x)+2y

y^((2))=2e^(2x)+2y^((1))

y^((3))=2^2e^(2x)+2y^((2))

...

y^((n))=2^(n-1)e^(2x)+2y^((n-1)).

I hope it is the answer you wanted!

Jim H
Apr 13, 2015

Thank you for providing a fun question to work on!

f(x) = xe^(2x)
f'(x) = e^(2x) + 2xe^(2x) = e^(2x)(2x+1)
f''(x) = 2e^(2x)(2x+1) +e^(2x)*2 = 2e^(2x)(2x+2)
f'''(x) = 4e^(2x)(2x+2) + 2e^(2x)*2 = 4e^(2x)(2x+3)

f^((4))(x) = 8e^(2x)(2x+3) + 8e^(2x) = 8e^(2x)(2x+4)

Looks like

f^((n))(x) = 2^(n-1)e^(2x)(2x+n)

It should be straightforward to prove by induction on n.

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