Results 1 to 7 of 7

Math Help - Finding Deriv when # is raised ^x

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    15

    Finding Deriv when # is raised ^x

    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

    This is all part of a larger problem which is:

    Find f'(x): (4^x*x^(1/3))/tanx

    Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Let 4^x = e^{ln(4^x)} = e^{xln(4)}. Thus the derivative, using the chain rule equals: e^{xln(4)}*ln(4). From the first equality, we know this equals 4^x*ln(4). The TI-89 says 4^x*2ln(2) because ln(4)=ln(2^2)=2ln(2).

    In general, \frac{d}{\,dx}c^x = ln(c)*c^x, where c is a constant.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by mattc View Post
    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

    This is all part of a larger problem which is:

    Find f'(x): (4^x*x^(1/3))/tanx

    Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?
    Let y = 4^x so that ln(y) = x\ln{4} = 2x\ln{(2)}

    Differentiating:
    <br />
(\frac{1}{y})(\frac{dy}{dx}) = 2\ln{(2)}

    multiply by y:

    \frac{dy}{dx} = 2\ln{(2)} y

    Recall that y= 4^x so we get

    \frac{dy}{dx} = 2\ln{(2)}*4^x

    In general: \frac{d}{dx}(a^x) = a\ln{(a^x)}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by mattc View Post
    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.
    The rule is

    \frac d{dx}\left[a^x\right]=(\ln a)a^x.

    To see why this is so, you may rewrite a^x as e^{x\ln a} and apply the more familiar rule for differentiating e^x.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    y = a^x<br />

    \ln{y} = \ln{a^x}

    \ln{y} = x \cdot \ln{a}

    \frac{d}{dx}[\ln{y} = x \cdot \ln{a}]

    \frac{y'}{y} = \ln{a}

    y' = \ln{a} \cdot y = \ln{a} \cdot a^x
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    With regard to your second question: \frac{d}{dx}\frac{4^xx^{1/3}}{tan(x)}, you want to use the product and quotient rules.

    \frac{d}{dx} = \frac{tan(x)*\frac{d}{dx}[4^xx^{1/3}]-4^xx^{1/3}*\frac{d}{dx}[tan(x)]}{tan^2(x)}

    \frac{d}{dx}[4^xx^{1/3}] = 4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x

    \frac{d}{dx}[tan(x)] = sec^2(x)

    Plugging this all yields: \frac{d}{dx} = \frac{tan(x)*(4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x)-4^xx^{1/3}*sec^2(x)}{tan^2(x)}

    I'll leave the simplification to you.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2009
    Posts
    15
    wow for some reason I had to keep staring at it but it sunk in.

    Thanks to all!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a formula for a matrix raised to the nth power
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 5th 2009, 04:03 PM
  2. Replies: 5
    Last Post: August 4th 2009, 01:28 PM
  3. log deriv.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 11th 2007, 08:40 AM
  4. deriv. of ln
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 8th 2007, 10:45 PM
  5. def. of deriv
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 7th 2007, 12:58 PM

Search Tags


/mathhelpforum @mathhelpforum