Question

Use mathematical induction to prove that...

Answer 1

\[f^{(n)}(x)=(2^nx+n2^{n-1})e^{2x}\]for all\[n \in \mathbb{Z}^+\], where \[f^{(n)}(x)\]represents the nth derivative of f(x).

Answer 2

Oh wait, \[f(x)=xe^{2x}\]

Answer 3

Base case: the derivative of \(xe^{2x}\) is \(e^{2x}+2xe^{2x}\), which is equivalent to \((2x+1)e^{2x}\).

Induction step: Show that if the \(n^{\text{th}}\) derivative of \(f(x)\) is \((2^nx+n2^{n-1})e^{2x}\), then the \((n+1)^{\text{th}}\) derivative is \((2^{n+1}x+(n+1)2^{n})e^{2x}\). Show this by deriving \((2^nx+n2^{n-1})e^{2x}\) with respect to \(x\).

Answer 4

I tried that, but I'm stuck.\[f^{(k+1)}(x)=\frac{d(f^k(x))}{dx}=\frac{d[(2^kx+k2^{k+1})e^{2x}]}{dx}\]Then I used the product rule\[\frac{d[(2^kx+k2^{k+1})e^{2x}]}{dx}=(2^k)e^{2x}+2e^{2x}(2^kx+k2^{k+1})\]\[\frac{dy}{dx}=e^{2x}[2^k+2(2^kx+k2^{k+1})]\]\[\frac{dy}{dx}=e^{2x}[2^k+2^{k+1}x+k2^{k+2})\]Now I'm stuck :(

Answer 5

you should be taking the derivative of \[(2^kx+k2^{k-1})e^{2x}\]but you have k+1 as the exponent.

Answer 6

if you make that correction the answer comes out the way its supposed to I believe.

Answer 7

Holy cow, you're right!! Oh was it really that simple? Aah how am I gonna pass if I make such stupid mistakes :(( Well thank you so much! Yep, now I got it :D