Second derivative of an exponential function

[Calculus :(]

Homework Statement

I have the derived function:
f'(x) = [1/(1+kx)^2]e^[x/(1+kx)]
k is a positive constant

Homework Equations

I need to find the second derivative, which I thought was just the derivative of the exponent multiplied by the coefficient (as you find the first derivative), but the answers say:

f''(x) = [1/(1+kx)^4]e^[x/(1+kx)] + e^[x/(1+kx)] . [-2k/(1+kx)^3]

The Attempt at a Solution

All I managed to get was:
f''(x) = [1/(1+kx)^4]e^[x/(1+kx)]

 

[Ray Vickson]

Homework Statement

I have the derived function:
f'(x) = [1/(1+kx)^2]e^[x/(1+kx)]
k is a positive constant

Homework Equations

I need to find the second derivative, which I thought was just the derivative of the exponent multiplied by the coefficient (as you find the first derivative), but the answers say:

f''(x) = [1/(1+kx)^4]e^[x/(1+kx)] + e^[x/(1+kx)] . [-2k/(1+kx)^3]

The Attempt at a Solution

All I managed to get was:
f''(x) = [1/(1+kx)^4]e^[x/(1+kx)]

You say "I thought was just the derivative of the exponent multiplied by the coefficient (as you find the first derivative)...", but THAT is not true: it is true ONLY if the "coefficient" of the exponential is a constant. If the coefficient itself is a function of x it will not work. Go back and review some basic properties of derivatives.

 

[Mandelbroth]

Homework Statement

I have the derived function:
f'(x) = [1/(1+kx)^2]e^[x/(1+kx)]
k is a positive constant

Homework Equations

I need to find the second derivative, which I thought was just the derivative of the exponent multiplied by the coefficient (as you find the first derivative), but the answers say:

f''(x) = [1/(1+kx)^4]e^[x/(1+kx)] + e^[x/(1+kx)] . [-2k/(1+kx)^3]

$\frac{df}{dx}=\frac{e^{\frac{x}{1+kx}}}{(1+kx)^2}$. You have the quotient of two functions of $x$. You can't just take the derivative of the exponential term. What do you do when you have a product or quotient of functions?