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Math Help - Differentiating Constant To Power x

  1. #1
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    Differentiating Constant To Power x

    How would you differentiate something like:

    y = 2^x With respect to x?
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    \frac{d}{dx}a^x = a^x \ln a
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    How about this question. Can someone do it:

    Using substitution of y=2^x, solve \int^1_0 \frac{2^x}{(2^x + 1)^2} \ \mathrm{d}x
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Air View Post
    How about this question. Can someone do it:

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

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

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

    The bounds are now from 2 to 3.

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

    It's easy from here..
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  5. #5
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    Quote Originally Posted by wingless View Post
    A better substitution is u = 2^x + 1.

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

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

    The bounds are now from 2 to 3.

    The integral becomes \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?
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    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:

    \frac{1}{\ln 2} \int_1^2 \frac{du}{(u + 1)^2} = \frac{1}{\ln 2} \int_1^2 (u + 1)^{-2} \, du.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Only a very minor change to what Wingless did is required:

    \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?
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  8. #8
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    Quote Originally Posted by Air View Post
    What rule would I use to integrate? Just reverse chain rule?
    Standard form ( n \neq -1): \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 .....
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  9. #9
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    Also, how do you differentiate:

    y=2^{x-1}.

    Is it the same rule?
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  10. #10
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    Hello,

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

    y=2^{x-1}.

    Is it the same rule?
    Yes it is.

    Actually, you can see it this way :

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

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

    Here, 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 2^{x-1}=\frac 12 \cdot 2^x
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