taking the derivative of exponential funcitons?

Question

f(x) = e^x/e^x + 1

I use the quotient rule which gets me

(e^x + 1)(e^x)-(e^x)(e^x)/(e^x + 1)^2

I have this gut feeling I did something wrong. Any ideas. What am I missing.

Answer 1

f(x) = e^x/(e^x + 1)

Taking the derivative (with the use of the quotient rule)
f'(x) = [(e^x + 1)(e^x) - (e^x)(e^x)]/(e^x + 1)²

Here is what you're missing.
Of course, you may simplify by distributing the monomials.
f'(x) = (e^2x + e^x - e^2x)/(e^x + 1)²

Combining like terms.
f'(x) = e^x/(e^x + 1)²

^_^
Hope this helps^_^
^_^

Answer 2

You did something wrong!
As you have written the question it reads as:-
f(x) = e^x / e^x + 1 = 1 + 1 = 2.
So in answer to your question , you are missing brackets!

Have now to assume that you mean:-
f(x) = e^x / (e^x + 1)
f `(x) = [ (e^(x + 1)).e^x - e^x.e^x ] / (e^x + 1)²
f `(x) = [ e^(2x) + e^x - e^(2x) ] / (e^x + 1)²
f `(x) = e^x / (e^x + 1)²

Answer 3

derivative of (af(x) +b)/(cf(x)+d)
= (ad - bc)f ' (x)/ (cf(x)+d)^2
This is a quick way to differntiate when you have same function on numerator and denominator. In your example a = c = d = 1 and b = 0
hence ad-bc = 1
derivative = e^x/(e^x +1) ^2

Another example: (3sinx + 1)/(2sinx-1)
a=3, b=1, c=2, d= -1
ad-bc = -5
derivative = -5cosx/(2sinx-1)^2

Answer 4

If you mean

f(x)=e^x/(e^x +1)

use the u/v rule.
=>g(x)=u/v

=>g'(x)=(vu' - uv')/v^2

u=e^x, v=e^x +1

=>[(e^x +1)e^x - e^x(e^x)]/(1+e^x)^2


=>e^x(1+e^x - e^x)/(1+e^x)^2

=>e^x/(1+e^x)^2

Answer 5

f'(x) = e^x/(e^x + 1)-e^(2x)/(e^x +1)^2
= e^x/(e^x +1)^2
=e^x(e^x+1)^-2

Answer 6

No you are doing it right..... the final answer will be

e^x/(e^x+1)^2

Answer 7

f(x) = e^x/(e^x + 1)
f'(x) = e^x/(e^x + 1)-e^(2x)/(e^x +1)^2 = e^x/(e^x +1)^2
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You did right. You can go one more step to simplify it.