I'm lost on the Quotient Rule.?

Question

(4x − 2) / (x^2 + 1)

Answer 1

The quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist. For example, if we want to find the derivatives of
f(x) = g(x)/h(x), we can use the quotient rule.

Here is simple way to remember the quotient rule:

df/dx = [low (de high)-(high) (de low)]/(low)^2

"de" stands for derivaties.
For example, let's take the derivatives of f(x) = g(x)/h(x). In this case, low = h(x), high = g(x)
Thus, d(f(x))/dx = [h(x) de(g(x))-g(x) de(h(x))/(h(x))^2

Let's do a simple example. Let's say we want to find the derivative of f(x) = (x+3)/(x-1).
Derivative of (x+3) = 1
Derivative of (x-1) = 1
Thus df(x)/dx = [(x-1)(1) - (x+3)(1)/(x-1)^2] which can be simplified to df(x)/dx = [-4/(x-1)^2]

Back to your question:
Derivative of (4x-2) = 4
Derivative of (x^2+1) = 2x
Thus using the quotient rule:
df/dx = [(x^2+1)(4)-((4x-2)(2x)/(x^2+1)^2
df/dx = [4x^2+4-8x^2+4x)/(x^2+1)^2]
df/dx = [x^2+4x+4)/(x^2+1)^2]
df/dx = (-4x^2+4x+4)/(X^4 + 2x^2 + 1)

Answer 2

The quotient rule is used to find the derivative of the above equation.

Just remember this:

Bottom times the derivative of the top, minus, top times the derivative of the bottom, ALL OVER the bottom squared.

So lets follow this sentence along:

Bottom = x^2+1
times
the derivative of the top = 4 (remember the derivative of 4x^1-2, take the exponent (the power) and multiply it by the number in front....this gives us 4 then reduce the exponent by 1...so 1 - 1 = 0 so what we have is 1*4x^0, but anything to the 0 power is 1. So this leaves us with 1*4 = 4) the 2 doesn't show up because the derivative of a constant is 0

So bottom times the derivative of the top is (x^2+1)*4 = 4x^2+4

continuing on

MINUS

Top= 4x-2
times the derivative of the bottom...take the exponent, multiply it out in front, reduce the exponent by 1....and, as before, the derivative of the constant, 1, is zero, so you have 2*1*x^1 or 2x

So the top times the derivative of the bottom is (4x-2)2x = 8x^2-4x

ALL OVER

the bottom squared = (x^2+1)^2 or expanding it out is (x^2+1)(x^2+1). Use the distributive rule...first outter inner last gives:

X^4 + 2x^2 + 1

So now take all of the parts and put them together

(Top*Der of Bottom - Bottom * der of top)/Bottom^2

Substitute in the values we have:

(4x^2+4 - (8x^2-4x))/(X^4 + 2x^2 + 1)

This is a good answer, but it can be reduced, look at the stuff in the numerator:

4x^2+4-8x^2+4x, combine terms with identical exponents:

-4x^2+4x+4

Now put this back on top of the denominator:

(-4x^2+4x+4)/(X^4 + 2x^2 + 1)

Hope this helps:)

Answer 3

If you don't like the quotient rule, write it as a product, and use the product rule!

(4x - 2)(x^2 + 1)^(-1)

Answer 4

A good mnemonic to help you with the quotient rule is this. Think of the fraction as Hi/Ho. Then, by the quotient rule we have:

d/dx(Hi/Ho) = [Ho d(Hi) - Hi d(ho)]/HoHo

I know it sounds silly, but it is a way to remember it.

In your case, Hi = (4x-2) and Ho = (x^2 + 1).

Using this rule, you should get:

-4(x^2 - x -1)/(x^2+1)^2

Let me know if you don't understand.

Answer 5

Product rule: duv == vdu + udv
Exponent Rule: du^n == ndu^(n-1)
Inverse Rule: du^(-1) == -(1/u^2)du
Quotient Rule: d(u/v) == d(u*v^(-1))
== u*(-1/v^2)dv + v^(-1)du
== (-u/v^2)dv +(1/v)du

Thus

(d/dx)((4x-2)/(x^2+1))

== (-(4x-2)/(x^2+1)^2)2x + (1/(x^2+1))4

== (-8x^2+4x)/(x^2+1)^2 + 4/(x^2+1)

== (-8x^2+4x+4x^2+4)/(x^2+1)^2

== 4(1+x-x^2)/(x^2+1)^2

Whew!

(Hope that's right -- doublecheck!!)

Answer 6

I agree with starman, the easiest way to remember it is a little song thingy, similar to his.

Low = denominator
High = numerator

"Low d-high minus high d-low, square the bottom and away we go"

d-ing something is taking its derivative.

So...

Low (x^2+1)
d-high (4)
minus -
High (4x-2)
d-low (2x)
Square the bottom (the denominator, that is) (x^2 + 1)^2
And away we go!

___4(x^2+1)_-_2x(4x-2)____
(x^2+1)^2

That is your answer. Simplify. (pretend those underscores are the fraction bar -____-)

Answer 7

(4x - 2)/(x^2 + 1)

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

f(x) = (4x - 2)
g(x) = (x^2 + 1)

f'(x) = (((4x - 2)'(x^2 + 1)) - ((4x - 2)(x^2 + 1)')))/((x^2 + 1)^2)
f'(x) = (4(x^2 + 1) - ((2x)(4x - 2))/((x^2 + 1)^2)
f'(x) = (4x^2 + 4 - (8x^2 - 4x))/((x^2 + 1)^2)
f'(x) = (4x^2 + 4 - 8x^2 + 4x)/((x^2 + 1)^2)
f'(x) = (-4x^2 + 4x + 4)/((x^2 + 1)(x^2 + 1))
f'(x) = (-4x^2 + 4x + 4)/(x^4 + x^2 + x^2 + 1)
f'(x) = (-4x^2 + 4x + 4)/(x^4 + 2x^2 + 1)

ANS : f'(x) = (-4x^2 + 4x + 4)/(x^4 + 2x^2 + 1)

Answer 8

quotient rule:

d(f/g)=(gf'-fg')/g^2

Answer 9

(denominator * derivative of numerator) - ( numerator * derivative of denominator)
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denominator square