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
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"
Now to your problem
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]