derivative of x/(1+x)?

Question

what is the derivative of x/(1+x)? it's been a while since calculus and i need this for my econ class.

Answer 1

Using the quotient rule:

d(x/(1+x))/dx = (d(x)/dx (1+x) - d(1+x)/dx x)/(1+x)² = (1+x-x)/(1+x)² = 1/(1+x)²

Alternatively, simplify it first by noting that x/(1+x) = 1-1/(1+x), then use the power and chain rules.

Answer 2

Take this as x * (1+x)^-1, then differentiate it as the product of two terms. So it's the derivative of the first times the second, plus the derivative of the second times the first.

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

Answer 3

Given function is H(x) = f(x) / g(x) = x / (1+x)

To find the derivative of this function we have to use the quotient rule.
That is H ' (x) = [ g(x) * f ' (x) - f(x) * g ' (x) ] / [ g(x) ] ^ 2

Here, f(x) = x ,
so, f ' (x) = 1 and g(x) = (1+x) implies g ' (x) = 1
Plug the values in the quotient rule,
we have,
H ' (x) = [ ( 1 + x ) * 1 - x * 1 ] / (1+x) ^ 2

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

= 1 / (1+x) ^ 2

This is the required derivative of the given function.

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Answer 4

in any fraction the derivative goes like this:
derivative of the numerator times the denomenator ,minus the derivative of the denmenator times the numerator,and all that is divided by the square the denomenator.
so the answer :
[(1+x)-(x)]/(1+x)^2 =1/(1+x)^2

Answer 5

Rewrite this a x * (1+x)^-1

Now, use the product rule. That is defined as u'v+v'u
In this case, let u=x and v=(1+x)^-1

So, u'v = 1*(1+x)^-1 = (1+x)^-1
v'u = -x*(1+x)^-2 * 1 =-x*(1+x)^2

So, the derivative is : (1+x)^-1 - x(1+x)^-2

or 1/(1+x) - x/(1+x)^2
Use (1+x)^2 as the least common denominator, and mulitply the first term by (1+x)/(1+x), you get:

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

Answer 6

FOIL the numerator and FOIL the denominator to avoid multiple use of the product rule within the quotient rule (x^2 + 4x + 3) / (x^2 - 4x + 3) now do quotient rule only: [ (2x+4) (x^2 - 4x + 3) - (2x-4) ((x^2 +4x + 3) ] / (x^2 - 4x + 3)^2 which does simplify: numerator: =2x^3 - 8x^2 + 6x + 4x^2 - 16x + 12 - ( 2x^3 + 8x^2+6x - 4x^2 - 16x - 12) =2x^3 - 8x^2 + 6x + 4x^2 - 16x + 12 - 2x^3 - 8x^2 - 6x + 4x^2 + 16x + 12 = - 8 x^2 + 24 ANSWER: [ -8x^2 + 24 ] / (x^2 - 4x + 3)^2

Answer 7

Short way:
x/(1+x) = (1+x-1)/(1+x) = 1 - 1/(1+x)
[x/(1+x)]' = 1/(1+x)^2

Answer 8

Quotient rule:
d/dx(u/v) = (u'v - uv')/v^2 =
[(1 + x) - x]/(1 + x)^2 = 1/(1 + x)^2

Answer 9

Using quotient rule:-
f `(x) = (v.du/dx - u.dv/dx) / v²
where u = x , du/dx = 1
v = 1 + x , dv/dx = 1
v² = (1 + x)²
f `(x) = ((1 + x) - x) / (1 + x)²
f `(x) = 1 / (1 + x)²