Results 1 to 5 of 5

Math Help - Derivatives of Exponential Functions

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    16

    Derivatives of Exponential Functions

    Hi Guys, can you explain a simple rule when finding the derivatives on exponential functions please?

    Find the Deriviative:

    g(x)=x^2e^4x^3

    Use the product rule, then simplfy.

    g'(x)=(x^2)'(e^4x^3) + (x^2)(e^4x^3)'
    g'(x)= (2x)(e^4x^3) + (x^2)(e^4x^3)(12x^2)
    g'(x)=2xe^4x^3+12x^4e^4x^3

    I understand this completely, except for the 12X^2 in the second step. In this situation, why does the derivative of the exponential function e^4x^3 become part of the base and not e^12x^2?

    I think this may solve the many problems I'm having with natural logarithms and exponential functions, staring at the function doesn't work anymore

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    Quote Originally Posted by Butum View Post
    Hi Guys, can you explain a simple rule when finding the derivatives on exponential functions please?

    Find the Deriviative:

    g(x)=x^2e^4x^3

    Use the product rule, then simplfy.

    g'(x)=(x^2)'(e^4x^3) + (x^2)(e^4x^3)'
    g'(x)= (2x)(e^4x^3) + (x^2)(e^4x^3)(12x^2)
    g'(x)=2xe^4x^3+12x^4e^4x^3

    I understand this completely, except for the 12X^2 in the second step. In this situation, why does the derivative of the exponential function e^4x^3 become part of the base and not e^12x^2?

    I think this may solve the many problems I'm having with natural logarithms and exponential functions, staring at the function doesn't work anymore

    thanks!
    The 12x^2 comes from the chain rule.

    In general, the derivative of

    e^{f(x)}

    is

    e^{f(x)} f'(x).

    For more arbitrary base:

    a^{h(x)}

    derives to

    \ln a a^{h(x)} h'(x)

    Example:

    The derivative of

    2^{\cos x}

    is

    \ln{2}\text{  }2^{\cos x} (-\sin x)

    Good luck!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    16
    So we cannot fully use the chain rule when Euler's constant is a base? Is this because Euler's Constant is not considered a positive constant?

    So for example, we will never see:

    y=e^-(2x+5)
    y'=e^-(2x+5)Ln e(-2)


    I apologize if this is breaking the forum rule for posting two questions within one topic.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member apcalculus's Avatar
    Joined
    Apr 2009
    From
    Boston
    Posts
    293
    The chain rule will always be applied as long as there is a composition of functions.

    Remember that ln e = 1, so we don't write it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    16
    That is so obvious to me now, thank you so much.

    I'll use the chain rule and just clear the Ln e when simplifying so I do not confuse myself.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives of Exponential Functions
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 24th 2011, 12:35 PM
  2. Derivatives of exponential functions
    Posted in the Calculus Forum
    Replies: 13
    Last Post: November 29th 2009, 12:26 PM
  3. partial derivatives of exponential functions
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 2nd 2009, 02:34 PM
  4. derivatives exponential functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 5th 2009, 07:49 PM
  5. Replies: 4
    Last Post: February 10th 2009, 09:54 PM

Search Tags


/mathhelpforum @mathhelpforum