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Math Help - Derivative with exponential f(x) in denominator

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    Derivative with exponential f(x) in denominator

    I am trying to take a derivative of the function below with respect to X. The problem is X is both in the numerator and in the exponential function in the denominator. I don't know how to get the .4X-7.8 out of the exponential function and then combine with the 16X to simplify and take a derivative? I was trying to use ln() somehow but can't seem to figure it out. Thanks!
     f(x)=(X-5)(15)\frac{1}{1+e^{0.4X-7.8}}
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  2. #2
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    Quote Originally Posted by atljogger View Post
    I am trying to take a derivative of the function below with respect to X. The problem is X is both in the numerator and in the exponential function in the denominator. I don't know how to get the .4X-7.8 out of the exponential function and then combine with the 16X to simplify and take a derivative? I was trying to use ln() somehow but can't seem to figure it out. Thanks!
     f(x)=(X-5)(15)\frac{1}{1+e^{0.4X-7.8}}
    is your function

    f(x)=\frac{(x-5)15}{1+e^{0.4x-7.8}}?

    if so, then use the quotient rule as follows:

    For a function f(x) = \frac{g(x)}{h(x)}, the derivative f'(x) = \frac{[g'(x) \times h(x)] - [g(x) \times h'(x)]}{{h(x)}^{2}}

    so you have,

     g(x) = x-5 and

     h(x) = 1+e^{0.4x-7.8}

    f'(x) = 15 \times \frac{(1-0) \times (1+e^{0.4x-7.8}) - ((x-5)(0+ e^{0.4x-7.8} \times 0.4)}{(1+e^{0.4x-7.8})^2}

    f'(x) = 15 \times \frac{(1+e^{0.4x-7.8}) - ((x-5)(e^{0.4x-7.8} \times 0.4))}{(1+e^{0.4x-7.8})^2}
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    Thanks! However the X variable is still in the exponent of the exponential function. The 2nd part of this question is to find the value of X that maximizes the function - this is a profit function. So I am trying to take the derivative of f(x) and then solve for X. How do I get the 0.4X out of the exponent?
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    One bump on this thread. If anyone can tell me the general rule to get an equation out of the exponent for an exponential function, I can probably figure this equation out. Thanks again!
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