# Math Help - Deriviative of function

1. ## Deriviative of function

If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).

I'm not sure what d^2 is supposed to mean.. does it mean take the derivative and square it?

Well I did take the derivative.. did not square it yet. Don't know if this is right but gave it a shot:

(d^(2)F)/(dS^2)=-e^-s -6e^(-2s)+k(-e^(-s) -6e^(-2s) (-2))
=e^-s(-6e^-2s+k(12e^-2s-e^-s)

2. Originally Posted by skeske1234
If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).

I'm not sure what d^2 is supposed to mean.. does it mean take the derivative and square it?
No, it is the second derivative

3. Originally Posted by running-gag
No, it is the second derivative
Ok i did that: found the 2nd derivative of F

I did not end up with the answer the wanted me to prove............

(12e^(-2s)-e^(-s))[1-24e^(-2s)+e^(-s)]

What did I do wrong? please demonstrate, ty

4. F=k(e^(-s) -6e^(-2s))

dF/ds = k(-e^(-s)+12e^(-2s))

d²F/ds² = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)

5. Originally Posted by running-gag
F=k(e^(-s) -6e^(-2s))

dF/ds = k(-e^(-s)+12e^(-2s))

d²F/ds² = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)
for the first derivative, did you not use the product rule? does k = 0?

6. You don't need to use the product rule since k is a constant
The derivative of (kf), where k is a constant and f a function, is kf'

7. Originally Posted by running-gag
You don't need to use the product rule since k is a constant
The derivative of (kf), where k is a constant and f a function, is kf'
how do you know if k is a constant or not
and

d²F/ds² = k(e^(-s)-24e^(-2s)) = F - 18e^(-2s)

How did you get the last answer
F-18e^(-2s)

How did that pop up all of a sudden from your previous answer?....

8. Originally Posted by skeske1234
If a force F is defined by F=k(e^(-s) -6e^(-2s)), where S is the distance between two objects, then prove (d^(2)F)/(d(S)^2)=F-18ke^(-25).
$F = k(e^{-s} - 6e^{-2s})$

$\frac{dF}{ds} = k(-e^{-s} + 12e^{-2s})$

$\frac{d^2F}{ds^2} = k(e^{-s} - 24e^{-2s})$

$\frac{d^2F}{ds^2} = k(e^{-s} - 6e^{-2s} - 18e^{-2s})$

$\frac{d^2F}{ds^2} = \textcolor{red}{k(e^{-s} - 6e^{-2s})} - 18ke^{-2s}$

$\frac{d^2F}{ds^2} = \textcolor{red}{F} - 18ke^{-2s}$