Results 1 to 10 of 10

Thread: Differentiating Constant To Power x

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Differentiating Constant To Power x

    How would you differentiate something like:

    $\displaystyle y = 2^x$ With respect to $\displaystyle x$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    $\displaystyle \frac{d}{dx}a^x = a^x \ln a$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    How about this question. Can someone do it:

    Using substitution of $\displaystyle y=2^x$, solve $\displaystyle \int^1_0 \frac{2^x}{(2^x + 1)^2} \ \mathrm{d}x$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by Air View Post
    How about this question. Can someone do it:

    Using substitution of $\displaystyle y=2^x$, solve $\displaystyle \int^1_0 \frac{2^x}{(2^x + 1)^2} \ \mathrm{d}x$
    A better substitution is $\displaystyle u = 2^x + 1$.

    Then, $\displaystyle du = 2^x \ln 2~dx$.

    $\displaystyle 2^x ~dx=\frac{du}{\ln 2}$

    The bounds are now from 2 to 3.

    The integral becomes $\displaystyle \frac{1}{\ln 2} \int_2^3 \frac{du}{u^2}$

    It's easy from here..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by wingless View Post
    A better substitution is $\displaystyle u = 2^x + 1$.

    Then, $\displaystyle du = 2^x \ln 2~dx$.

    $\displaystyle 2^x ~dx=\frac{du}{\ln 2}$

    The bounds are now from 2 to 3.

    The integral becomes $\displaystyle \frac{1}{\ln 2} \int_2^3 \frac{du}{u^2}$

    It's easy from here..
    Yes, I would have done that too but they provided the substitution value. How would I go about with that?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Air View Post
    Yes, I would have done that too but they provided the substitution value. How would I go about with that?
    Only a very minor change to what Wingless did is required:

    $\displaystyle \frac{1}{\ln 2} \int_1^2 \frac{du}{(u + 1)^2} = \frac{1}{\ln 2} \int_1^2 (u + 1)^{-2} \, du$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by mr fantastic View Post
    Only a very minor change to what Wingless did is required:

    $\displaystyle \frac{1}{\ln 2} \int_1^2 \frac{du}{(u + 1)^2} = \frac{1}{\ln 2} \int_1^2 (u + 1)^{-2} \, du$.
    What rule would I use to integrate? Just reverse chain rule?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Air View Post
    What rule would I use to integrate? Just reverse chain rule?
    Standard form ($\displaystyle n \neq -1$): $\displaystyle \int (au + b)^n \, du = \frac{1}{a} \, \frac{1}{(n+1)} \, (au + b)^{n+1} (+ C) $.

    Or you could make another substitution: w = u + 1 .....
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Also, how do you differentiate:

    $\displaystyle y=2^{x-1}$.

    Is it the same rule?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by Air View Post
    Also, how do you differentiate:

    $\displaystyle y=2^{x-1}$.

    Is it the same rule?
    Yes it is.

    Actually, you can see it this way :

    $\displaystyle a^{f(x)}=e^{f(x) \ln a}$

    --> $\displaystyle \left(a^{f(x)}\right)'=\ln a \cdot f'(x) e^{f(x) \ln a}=f'(x) \ln a \cdot a^{f(x)}$

    Here, $\displaystyle f(x)=x-1$

    You could also find this result by applying the chain rule and using wingless's formula


    In another way, you can see that $\displaystyle 2^{x-1}=\frac 12 \cdot 2^x$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Aug 31st 2011, 12:27 AM
  2. Replies: 1
    Last Post: Mar 7th 2011, 03:26 AM
  3. Manipulation of the constant C that ocurres after differentiating
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Dec 10th 2009, 11:49 AM
  4. Replies: 5
    Last Post: Feb 29th 2008, 02:05 PM
  5. constant to the power of x???
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Dec 1st 2007, 02:36 PM

Search Tags


/mathhelpforum @mathhelpforum