Derivative with exponential f(x) in denominator

• Mar 24th 2010, 04:25 PM
atljogger
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!
$f(x)=(X-5)(15)\frac{1}{1+e^{0.4X-7.8}}$
• Mar 24th 2010, 04:37 PM
harish21
Quote:

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

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

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

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

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

$f'(x) = 15 \times \frac{(1+e^{0.4x-7.8}) - ((x-5)(e^{0.4x-7.8} \times 0.4))}{(1+e^{0.4x-7.8})^2}$
• Mar 24th 2010, 05:01 PM
atljogger
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?
• Mar 25th 2010, 05:00 AM
atljogger
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!