Question

the derivative of (x)/(x+1)

Answer 1

the derivative of 1/x = ln(x) right? would the derivative of x/x+1 be xln(x)?!

Answer 2

http://en.wikipedia.org/wiki/Quotient_rule

Answer 3

is it (x+1)-x / (x+1)^2 ?

Answer 4

so the derivative of 1/x is not lnx? haha

Answer 5

you can either use product rule or quient rule

assume the function is f(x)
assume the neumorator as g(x)
and the denominator as s(x)

g(x) = x g'(x) = 1
s(x) = (x+1) s'(x) = 1
Using quient rule

f'(x) = g'(x)s(x) - s'(x)g(x)/(s(x))^2

so yeah sub in the values and you have or you can use product rule
assume the neumorator as g(x)
and the denominator as s(x)

where f(x) = x(x+1)^(-1)

g(x) = x g'(x) = 1
s(x) = (x+1)^(-1) s'(x) = -(x+1)

you need to use chain rule to fin the derivative of s(x)

thus

f'(x) = g'(x)s(x) + s'(x)g(x)

Answer 6

do you know the product rule? you can look at this as x(x+1)^-1

Answer 7

so is what i put incorrect?

Answer 8

No itzmashy read what I wrote

Answer 9

i did haha, i'm not sure what i did wrong then

Answer 10

yes, but product rule is easier to remember
and the quotient rule is just a special case of the product rule

Answer 11

assume the function is f(x) assume the neumorator as g(x) and the denominator as s(x)

g(x) = x g'(x) = 1
s(x) = (x+1) s'(x) = 1

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

Answer 12

yeah I know philips :) I just have it memorized so I would use it in a case like this

Answer 13

(x) = x g'(x) = 1
s(x) = (x+1) s'(x) = 1
Using quient rule

f'(x) = g'(x)s(x) - s'(x)g(x)/(s(x))^2

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

i simplified to (x+1) - x/ (x+1)^2

how is that different from: is it (x+1)-x / (x+1)^2 ?

Answer 14

well first of all you need to put brackets around the neumorator but they are the same otherwise

Answer 15

so i did do it correctly? i thought you said it was wrong? i'm confused now

Answer 16

It is correct, there is just another method you can use on these questions involving chain rule if you know what that is?

Answer 17

using chain rule u get -x(x+1)^-2, right?

Answer 18

it would give you the same answer I explained it above but I was lazy and didnt explain how I got the derivative of (x+1)^(-1)

Answer 19

oh yeah you are right :) I made a mistake

Answer 20

wait.. i think i did something wrong there... lol

Answer 21

remember chain rule is
g(x) = outside function
d(x) = inside function

chain rule is

g'(d(x)) * d'(x)

Answer 22

d/dx x = 1

Answer 23

thus 1 * -(1+x)^(-2)

Answer 24

I mean you wouldn't get x because it differentiates to 1

Answer 25

so am i using product rule and chain rule, or just chain rule?
if its product rule then isn't it:

-x(x+1)^-2+(x+1)^-1?

and if its just chainrule then:
-x(x+1)^-2?

Answer 26

no stop I will explain product rule lol

Answer 27

i know what the product rule is, i'm just not understanding how you used it in this case

Answer 28

yeah Im not saying yu dont know what it is but i will show you how to apply it in this instance

Answer 29

so for product rule

f(x) = x(x+1)^(-1)

g(x) = x g'(x) = 1
s(x) = (x+1)^(-1) s'(x) = -(x+1)^(-2)

we have to use chain rule to find s'(x) so lets assume d(x) is the outside function and m(x) is the inside function of s(x)

so

d(x) = x^(-1) d'(x) = -x^(-2)
m(x) = x +1 m'(x) = 1 + 0

now chain rule is d'(m(x)) * m'(x)
so we have that
s'(x) = -(x+1)^(-2) * 1 = -(x+1)^(-2)

thus f'(x) = g'(x)s(x) + s'(x)g(x)

so the answer is 1(x+1)^(-1) + -(x+1)^(-2)(x)

so I just read your question wrong sorry my brain is litterally dead from exams

Answer 30

no more confusion now

Answer 31

ooh okay that makes sense now, thanks!