Second derivative of an exponential function

In summary, the second derivative of the given function is equal to [1/(1+kx)^4]e^[x/(1+kx)] + e^[x/(1+kx)] . [-2k/(1+kx)^3], and the process of finding it involves taking the derivative of the exponent multiplied by the quotient of the two functions.
  • #1
Calculus :(
1
0

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)]

 
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  • #2
Calculus :( said:

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.
 
  • #3
Calculus :( said:

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?
 

Related to Second derivative of an exponential function

1. What is the second derivative of an exponential function?

The second derivative of an exponential function is the rate of change of the first derivative. It measures how quickly the slope of the original exponential function is changing at any given point.

2. How do you find the second derivative of an exponential function?

To find the second derivative of an exponential function, you first need to take the derivative of the function using the rules of differentiation. Then, you take the derivative of that result to find the second derivative. This can be done using the chain rule or the quotient rule, depending on the form of the original function.

3. Why is the second derivative of an exponential function important?

The second derivative of an exponential function is important because it can provide information about the concavity and inflection points of the original function. It can also help in determining the maximum and minimum values of the function.

4. Can the second derivative of an exponential function be negative?

Yes, the second derivative of an exponential function can be negative. This indicates that the slope of the original function is decreasing at that point. It can also mean that the function is concave down at that point.

5. How does the second derivative of an exponential function relate to the original function?

The second derivative of an exponential function is related to the original function in that it describes the rate of change of the first derivative. It can also provide information about the shape and behavior of the original function, such as its concavity or inflection points.

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