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Math Help - e^x containing composite function

  1. #1
    Senior Member Paze's Avatar
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    e^x containing composite function

    I'm wondering...With e^x

    Why doesn't \ln x count as an "inside" function?

    Example:

    \frac{dy}{dx}e^{sin(x)}=e^{sin(x)}cos(x)

    Here we see an example of the chain rule. Derivative of the inside function times the derivative of the outside function...

    However, with this example:

    \frac{dy}{dx}e^{\ln 7x}=e^{\ln 7x}\cdot \ln 7

    There is no derivative of the inside and then outside. It seems to adhere to normal e^x differentiation rules such as:

    \frac{dy}{dx}e^{7x}=7\cdot e^{7x}

    Why is this? Thanks!
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  2. #2
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    Re: e^x containing composite function

    Quote Originally Posted by Paze View Post
    example:

    \frac{dy}{dx}e^{\ln 7x}=e^{\ln 7x}\cdot \ln 7

    There is no derivative of the inside and then outside. It seems to adhere to normal e^x differentiation rules such as:

    \frac{dy}{dx}e^{7x}=7\cdot e^{7x}
    I simply do not understand what you are on about.

    First e^{\ln 7x}=7x

    so \frac{dy}{dx}e^{\ln 7x}=7.
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    Senior Member Paze's Avatar
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    Re: e^x containing composite function

    Quote Originally Posted by Plato View Post
    I simply do not understand what you are on about.

    First e^{\ln 7x}=7x

    so \frac{dy}{dx}e^{\ln 7x}=7.
    Sorry. It's supposed to say e^{(\ln 7)\cdot x}

    Do you understand now?
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  4. #4
    Member agentmulder's Avatar
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    Re: e^x containing composite function

    That's because the derivative of u = sinx is more complicated than the derivative of u = cx for any constant c. in your example the constant is c = ln (7).

    Thanks from Paze
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    Senior Member Paze's Avatar
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    Re: e^x containing composite function

    Quote Originally Posted by agentmulder View Post
    That's because the derivative of u = sinx is more complicated than the derivative of u = cx for any constant c. in your example the constant is c = ln (7).

    So logarithms aren't functions?
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    Re: e^x containing composite function

    Quote Originally Posted by Paze View Post
    So logarithms aren't functions?
    Of course not; functions map a range of inputs to a range of outputs, so therefore require at least a single variable to do so. So for instance e^x is a function that maps x to e^x. e^5 is not a function, it always has constant value. Ln(7x) is a function, but ln(7) is not. The derivative (wrt to x) of e^(ln7x) is as has been pointed out, 7, but the derivate of e^((ln7)x) is ln(7)e^((ln7)x) since it is of the form e^kx where k=ln7 so you can see it does indeed follow normal differentiation laws.
    Thanks from Paze
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    Senior Member Paze's Avatar
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    Re: e^x containing composite function

    Quote Originally Posted by Nemesis10192 View Post
    Of course not; functions map a range of inputs to a range of outputs, so therefore require at least a single variable to do so. So for instance e^x is a function that maps x to e^x. e^5 is not a function, it always has constant value. Ln(7x) is a function, but ln(7) is not. The derivative (wrt to x) of e^(ln7x) is as has been pointed out, 7, but the derivate of e^((ln7)x) is ln(7)e^((ln7)x) since it is of the form e^kx where k=ln7 so you can see it does indeed follow normal differentiation laws.
    Right. Thanks!
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  8. #8
    Member agentmulder's Avatar
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    Re: e^x containing composite function

    Uuhhh... it depends on what we are talking about... is a single 1 dimensional number a function ? I don't know. In 2 dimensions f(x) = c is a function , it is a horizontal line that takes on all x values exactly once and every x value has only 1 y value and therefore passes the so called 'vertical line test'. Similarly f(x) = ln 7 is a function in 2 (or more) dimensions. For any real number c , one may decide to rewrite this as

     f(x) = cx^0

    In order to have a 'place' to sbstitute for x ... this works out fine for all real x EXCEPT x = 0 where we encounter the 'disagreeable'

     f(0) = c \cdot 0^0

    Personally, it doesn't bother me too much and i won't throw it away just because there is a problem at x = 0 , but that's me , others may disagree.

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