Hi, I have this diff. question which i keep getting wrong:

Find dy/dx of:

v=1/(1+e^-x)

This is my working:

v=1/1 + 1/(e^-x)=1+1/(e^x)=1+e^x

therefore dy/dx would equal :

e^x

could someone please show me where i'm going wrong...

Thank you

2. Originally Posted by Kris
Hi, I have this diff. question which i keep getting wrong:

Find dy/dx of:

v=1/(1+e^-x)

This is my working:

v=1/1 + 1/(e^-x)=1+1/(e^x)=1+e^x

therefore dy/dx would equal :

e^x

could someone please show me where i'm going wrong...

Thank you
I'm curious as to how you got to calculus 1 without knowing how to deal with fractions.

Firstly,
In general: a/(b + c) IS NOT EQUAL TO a/b + a/c

so 1/(1+e^-x) IS NOT THE SAME AS 1/1 + 1/e^-x

we can only split up a fraction in that way if we have a single term in the denominator.
for example, (a + b)/c = a/c + b/c

so first thing you need to do is review fractions.

Secondly,
dy/dx means "the derivative of y with respect to x" there is no y here. you wan dv/dx, "the derivative of v with respect to x"

v = 1/(1+e^-x) ............change this in terms of an exponent
=> v = (1 + e^-x)^-1 ......now we proceed by the Chain rule

Note: Chain Rule
[f(g(x))]' = f'(g(x))*g'(x)

So, v = (1 + e^-x)^-1
=> dv/dx = -1(1 + e^-x)^-2 *(1 + e^-x)'
.........= -(1 + e^-x)^-2 * (-e^-x)
.........= (e^-x)(1 + e^-x)^-2
.........= (e^-x)/[(1 + e^-x)^2]

3. thank you, that was just a stupid mistake about the fractions, wasn't thinking straight yesterday and just couldn't figure it!