f(x) = xe^(8x). Find the nth derivative of f.

I took d/dx
f'(x) = 8xe^(8x) + e^(8x)
f''(x) = 64xe^(8x) + 16e^(8x)
f'''(x) = 512xe^(8x) + 192e^(8x)

So, 8^(n)xe^(8x)+....
I don't see the pattern for the second term :(

Question

f(x) = xe^(8x). Find the nth derivative of f.

I took d/dx
f'(x) = 8xe^(8x) + e^(8x)
f''(x) = 64xe^(8x) + 16e^(8x)
f'''(x) = 512xe^(8x) + 192e^(8x)

So, 8^(n)xe^(8x)+....
I don't see the pattern for the second term :(

freckles · Answer

$$f^{(0)}=xe^{8x}$$
$$f^{(1)}=(1)e^{8x}+x \cdot 8 \cdot e^{8x}$$
$$f^{(2)}=8e^{8x}+8e^{8x}+x \cdot 64 \cdot e^{8x}$$
$$f^{(3)}=64e^{8x}+64e^{8x}+64e^{8x} +x 64(8)e^{8x}$$
$$f^{(n)}=8^{n-1} \cdot e^{8x} n+x8^n e^{8x}$$
let me check....
what is 
$$f^{(4)}$$ using what i wrote then also using what is 
$$(f^{(3)})'$$

freckles · Answer

$$f^{(4)}=3\cdot 64(8)e^{8x}+64(8)e^{8x}+x64(8)(8)e^{8x}=4 \cdot 8^3+x8^4e^{8x}$$

freckles · Answer

using the "formula for n" i just wrote i get....

freckles · Answer

$$f^{(4)}=8^3e^{8x} \cdot 4 +x8^4 e^{8x}$$

freckles · Answer

these look the same i made a type for f^(4) above i hope you see it

freckles · Answer

$$f^{(4)}=3\cdot 64(8)e^{8x}+64(8)e^{8x}+x64(8)(8)e^{8x}=4 \cdot 8^3 e^{8x}+x8^4e^{8x}$$

freckles · Answer

that is what it was suppose to say
and these guys are the "samseys"

anonymous · Answer

Oh I see it now, thanks

freckles · Answer

sometimes it can be tricky finding or spotting patterns

freckles · Answer

it probably would have been nice of me to write each my f^(i) 's has compactly as  i could