# Thread: Derivative with exponential f(x) in denominator

1. ## 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!
$\displaystyle f(x)=(X-5)(15)\frac{1}{1+e^{0.4X-7.8}}$

2. Originally Posted by atljogger
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!
$\displaystyle f(x)=(X-5)(15)\frac{1}{1+e^{0.4X-7.8}}$

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

if so, then use the quotient rule as follows:

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

so you have,

$\displaystyle g(x) = x-5$ and

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

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

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

3. 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?

4. 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!