What is the derivative of:
\(f(x)=\left(1-kx\right)ae^{-kx}\)
a and k are constants.
Using the product rule
f'(x) = f'(u)*f(v) + f(u)*f'(v)
f(u) = 1-kx
f'(u) = -k
f(v) = \(ae^{-kx}\)
f'(v) = \(-ake^{-kx}\)
\(f'\left(x\right)=-kae^{-kx}+\left(1-kx\right)\left(-ake^{-kx}\right)\)
\(f'\left(x\right)=kae^{-kx}-kae^{-kx}+k^2xae^{-kx}\)
\(f'\left(x\right)=k^2xae^{-kx}\)
I've messed up somewhere...
Okay simple mistake (forgot a - sign).
My answer however ended up being.
ake-kx(kx-2)
And the answer in the book is:
ae-kx(k2x-2k)
Is it standard practice to keep 'like' terms together to make it simplier to find 3rd derivatives/inflection points? (the original function is actually the first derivative, so here we were finding the 2nd derivative).
Hi Alan,
Yes I know these are the same, my question was more... would there be a specific reason to give the result as:
ae-kx(k2x-2k)
Both k2x-2k and kx-2 are suitable for finding the inflection point, so I'm just curious if there is some specific reason to list it that way (typically I just factor everything I can).
+33615 I don't think there's a strong reason for expressing the answer this way. You could have asked the same question if it had been expressed your way! Sometimes it is easier to do further derivatives if the results are given in one form rather than another, but in this case either form is easy to take further it seems to me. Certainly, there is no general rule I'm aware of that specifies what form such a derivative should take.
