Results 1 to 6 of 6

Math Help - Finding slope of a tangent to a function

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    16

    Finding slope of a tangent to a function

    Hi guys,

    Another question I'm confused with. Question is:

    Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.

    First I start with the chain rule:

    f(x)=2^(x^2+3x)
    f'(x)=2^(x^2+3x) ln2 (2x+3)

    Then I input the value of x=3 into the equation.

    f'(3)=2^(3^2+3(3)) ln2 (2(3)+3)
    f'(3)=2^(3^2+3(3)) ln2 (9)

    This is where I get stuck, if I complete the function I am left with:

    f'(3)=2^18ln2 (9)
    f'(3)= 1,635,229.37. Obviously this is not the correct answer.

    I'm stuck, again.
    Last edited by Butum; March 17th 2010 at 07:18 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,114
    Thanks
    988
    Quote Originally Posted by Butum View Post
    Hi guys,

    Another question I'm confused with. Question is:

    Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.

    First I start with the chain rule:

    f(x)=2^x^2+3x
    f'(x)=2^x^2+3x ln2 (2x+3)

    Then I input the value of x=3 into the equation.

    f'(3)=2^3^2+3(3) ln2 (2(3)+3)
    f'(3)=2^3^2+3(3) ln2 (9)

    This is where I get stuck, if I complete the function I am left with:

    f'(3)=2^18ln2 (9)
    f'(3)= 1,635,229.37. Obviously this is not the correct answer.

    I'm stuck, again.
    first, these questions should be posted in the calculus forum, not pre-calculus.

    second, your notation leaves much to be desired. use parentheses to make your expressions clear.

    from your work, I assume your function is

    f(x) = 2^(x^2+3x) rather than f(x) = 2^(x^2) + 3x

    f'(x) = 2^(x^2+3x) * (2x+3) * ln(2)

    f'(3) = 2^(18) * (9) *ln(2) ... which, yes, is a very large number.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    16
    You are correct with your assumptions, I've edited the original post.

    Can this thread be moved to the proper section?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    16
    I still have not been able to solve this problem, perhaps I'm using the wrong rule?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,114
    Thanks
    988
    Quote Originally Posted by Butum View Post
    I still have not been able to solve this problem, perhaps I'm using the wrong rule?
    f'(3) = 2^{18} \cdot 9 \cdot \ln(2) = 1635339.37051
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    16
    Thanks skeeter, that's one of the two answers that I have been able to come up with.

    Is it common to see numbers that large when using the chain rule on polynomial expressions in the exponent? At first glance I feel like that is a very wrong answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: September 11th 2011, 05:53 PM
  2. Replies: 6
    Last Post: January 12th 2011, 03:38 PM
  3. Finding the slope of the tangent
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 22nd 2010, 04:18 PM
  4. finding the slope of the tangent line
    Posted in the Calculus Forum
    Replies: 9
    Last Post: November 30th 2009, 03:17 AM
  5. Replies: 2
    Last Post: May 15th 2008, 04:28 AM

Search Tags


/mathhelpforum @mathhelpforum