Results 1 to 12 of 12

Math Help - [SOLVED] Derivative

  1. #1
    Member ronaldo_07's Avatar
    Joined
    Nov 2008
    Posts
    175

    [SOLVED] Derivative

    For every natural number n, the derivative of x^n is

    (x^n)\prime=nx^{n-1}

    using the product rule: (f g)\prime =f \prime{g}+ f {g}\prime.
    ----------------------------------------------------------------------
    I have worked out  u=n, v=x^{n-1}, \frac{du}{dx}=1, \frac{dv}{dx}=x^{n-2}

    Therefore (f g)\prime =nx^{n-2}+x^{n-1}

    Is this correct so far?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,685
    Thanks
    446
    Quote Originally Posted by ronaldo_07 View Post
    (x^n)\prime=nx^{n-1}

    using the product rule: (f g)\prime =f \prime{g}+ f {g}\prime.
    ----------------------------------------------------------------------
    I have worked out  u=n, v=x^{n-1}, \frac{du}{dx}=1, \frac{dv}{dx}=x^{n-2}

    Therefore (f g)\prime =nx^{n-2}+x^{n-1}

    Is this correct so far?
    if n is a constant ... no, it is not correct.

    sure you didn't mean for u = x^n instead of just n ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member ronaldo_07's Avatar
    Joined
    Nov 2008
    Posts
    175
    I am not sure how to solve this them please help, I have added the 1st line of the question now

    Do I make x^{n-1}=(x^n)(n^-1)

    so that u=x^n and   v=x^{-1}
    Last edited by ronaldo_07; January 16th 2009 at 05:49 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1
    Using the product rule is useless. n is a constant and its derivative is 0 since the slope of a constant (e.g. y=1) is 0.
    If you want to prove that  (x^n)\prime=nx^{n-1} then I fear that you have no choice but to compute
     \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{(x + \Delta x)^n - x^n}{\Delta x} = nx^{n-1} I guess that you can use the intuitive notion of a limit to solve this.
    Expand the whole thing, simplify (substract  x^n ), divide by  \Delta x and compute the limit (everything goes to 0 except one term).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member ronaldo_07's Avatar
    Joined
    Nov 2008
    Posts
    175
    The question specifically asks to use the product rule as follows:

    Use induction to prove the statement:

    For every natural number n, the derivative of x^n is

    (x^n)\prime=nx^{n-1}

    You will need to use the product rule: (f g)\prime =f \prime{g}+ f {g}\prime.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ronaldo_07 View Post
    The question specifically asks to use the product rule as follows:

    Use induction to prove the statement:

    For every natural number n, the derivative of x^n is

    (x^n)\prime=nx^{n-1}

    You will need to use the product rule: (f g)\prime =f \prime{g}+ f {g}\prime.
    Let P(n) be the statement we want to prove.

    We proceed by induction on n, where n \in \mathbb{N} = \{ 0, 1 , 2, 3, ... \}

    Clearly P(0) is true, since (x^0)' = 0 = 0x^{0-1} ....(of course, x \ne 0. if you want x to be anything it wants to be, then take the natural numbers greater than or equal to 1 in your statement)

    Assume P(n) is true, we show P(n + 1) is true.

    Now, x^{n + 1} = x \cdot x^n = \cdots

    basically, you want to apply the product rule, and simplify to get (n + 1)x^n, which would be P(n + 1)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member ronaldo_07's Avatar
    Joined
    Nov 2008
    Posts
    175
    Quote Originally Posted by Jhevon View Post
    Let P(n) be the statement we want to prove.

    We proceed by induction on n, where n \in \mathbb{N} = \{ 0, 1 , 2, 3, ... \}

    Clearly P(0) is true, since (x^0)' = 0 = 0x^{0-1} ....(of course, x \ne 0. if you want x to be anything it wants to be, then take the natural numbers greater than or equal to 1 in your statement)

    Assume P(n) is true, we show P(n + 1) is true.

    Now, x^{n + 1} = x \cdot x^n = \cdots

    basically, you want to apply the product rule, and simplify to get (n + 1)x^n, which would be P(n + 1)
    Basically I cant use product rule until i have applied N+1 in to the derivative right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ronaldo_07 View Post
    Basically I cant use product rule until i have applied N+1 in to the derivative right?
    i do not understand what you mean. use the product rule to differentiate x \cdot x^n. we can differentiate each of those terms because P(n) is true. so we can validly apply the product rule here

    although, i am thinking of starting with P(1) instead of P(0). what is the definition of natural numbers that you are using. including zero has some annoying cases that we may have to deal with explicitly.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member ronaldo_07's Avatar
    Joined
    Nov 2008
    Posts
    175
    I missread I got it thanks, We do Include 0 as a natural number in our study.

    Thanks for the help
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ronaldo_07 View Post
    I missread I got it thanks, We do Include 0 as a natural number in our study.

    Thanks for the help
    good.

    in that case, you should explicitly show that x follows the rules. that is, P(1) is true.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,685
    Thanks
    446
    Quote Originally Posted by ronaldo_07 View Post
    The question specifically asks to use the product rule as follows:

    Use induction to prove the statement:

    For every natural number n, the derivative of x^n is

    (x^n)\prime=nx^{n-1}

    You will need to use the product rule: (f g)\prime =f \prime{g}+ f {g}\prime.
    it never ceases to amaze me the results one can get when they post the original question in its entirety.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] second derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 24th 2009, 04:59 PM
  2. [SOLVED] derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 21st 2009, 01:55 AM
  3. [SOLVED] Derivative of log.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 3rd 2009, 09:09 PM
  4. [SOLVED] Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 22nd 2008, 10:54 PM
  5. [SOLVED] [SOLVED] help with a derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 18th 2008, 09:34 PM

Search Tags


/mathhelpforum @mathhelpforum