# Derive Quotient Rule: Find the Mistake

• A_Munk3y
In summary: I just realized that I didn't realize that x^2+2x+1 was a Quotient and I was trying to do the Product Rule. I feel a little dumb now.In summary, the student was trying to use the Product Rule to solve a derivative problem, but got confused because the steps were not explained clearly. The student was then helped to solve the problem using the Quotient Rule.
A_Munk3y

## Homework Statement

derive f(x)=x2+2x+1/ x2-3x+2

## The Attempt at a Solution

(x2-3x+2)(d/dx)(x2+2x+1)-(x2+2x+1)(d/dx)(x2-3x+2)/(x2-3x+2)2
=(x2-3x+2)(2x+2)-(x2+2x+1)(2x-3)
=(2x3-4x2-2x+4)-(2x3+x2-4x-3)
=5x2+2x+7/(x2-3x+2)2

that's what I'm getting... but according to the online derivative calculator, it's 5x2-2x-7/(x2-3x+2)2

what am i doing wrong? i keep checking the simplification and i keep getting the same answer

A_Munk3y said:

## Homework Statement

derive f(x)=x2+2x+1/ x2-3x+2

## The Attempt at a Solution

(x2-3x+2)(d/dx)(x2+2x+1)-(x2+2x+1)(d/dx)(x2-3x+2)/(x2-3x+2)2
=(x2-3x+2)(2x+2)-(x2+2x+1)(2x-3)
=(2x3-4x2-2x+4)-(2x3+x2-4x-3)
=5x2+2x+7/(x2-3x+2)2

that's what I'm getting... but according to the online derivative calculator, it's 5x2-2x-7/(x2-3x+2)2

what am i doing wrong? i keep checking the simplification and i keep getting the same answer

Would you be allowed to use the Product Rule to prove the Quotient Rule?Edit:

Product Rule:
f'(x) = [g(x)*u(x)]d/dx
= g(x)*u(x)' + u(x)*g(x)'

Quotient Rule:
f'(x) = [g(x)/u(x)]d/dx
= [g(x)u(x)' - u(x)g(x)] / u(x)^2

Last edited:
I don't know lol... how would i do that anyways?

A_Munk3y said:
I don't know lol... how would i do that anyways?

f(x)=x^2+2x+1/ x^2-3x+2

So using the Product Rule, you could just let your
g(x) = x^2 +2x +1, and your u(x) = (x^2 -3x +2)^-1

Try doing that, and keep the negative exponent when you differentiate using the Product Rule. Let me know if you get the correct answer that way :)

My result is
$$(-5x^2+2x+7)/(x^2-3x+2)^2$$

You missed a sign before the 5x^2

your calculator is wrong ? you missed a sign copying the calculator ??

Quinzio said:
My result is
$$(-5x^2+2x+7)/(x^2-3x+2)$$

You missed a sign before the 5x^2

your calculator is wrong ? you missed a sign copying the calculator ??

f'x = -(5*x^2-2*x-7)/(x^4-6*x^3+13*x^2-12*x+4) is what I got. When using the Quotient Rule, don't you have to square the bottom Quinzio?

silvermane said:
f'x = -(5*x^2-2*x-7)/(x^4-6*x^3+13*x^2-12*x+4) is what I got. When using the Quotient Rule, don't you have to square the bottom Quinzio?

OK ok I have the squared. Mess up with latex

Sorry, I am a little confused here :P just started derivatives.
Anyways, before i go any further, is this what i am supposed to do for the product rule?

(x2+2x+1)(d/dx)(x2-3x+2)-1+(x2-3x+2)-1)(d/dx)(x2+2x+1)
edit:
Ok, i think i got it using the quotient rule...
i got
-5x2+2x+7/(x5x2-3x+2)5x2
or if simplify the bottom then
-5x2+2x+7/x4-6x3+132-12x+4

but still, i would like to try and check it using the product rule

Last edited:
A_Munk3y said:
Sorry, I am a little confused here :P just started derivatives.
Anyways, before i go any further, is this what i am supposed to do for the product rule?

(x2+2x+1)(d/dx)(x2-3x+2)-1+(x2-3x+2)-1)(d/dx)(x2+2x+1)

edit:
Ok, i think i got it using the quotient rule...
i got
-5x2+2x+7/(x5x2-3x+2)5x2
or if simplify the bottom then
-5x2+2x+7/x4-6x3+132-12x+4

but still, i would like to try and check it using the product rule

To do this using the Product Rule, you have to know a secret. I didn't realize that you were just learning the rules because you said that you were "deriving" the Quotient Rule. That created a lot of misunderstandings between what you were asking and what I thought you wanted. Because of this, I thought that you knew more than you actually did, and told you the alternative method for using the Quotient Rule.

When you learn about the chain rule, you can apply the Product Rule to any Quotient such as the one you were asking for help on.

The Chain Rule is:
f(x)' = g(u(x)) = g(u(x))' * u(x)

An example would be:
f(x) = (x^2+2x+1)^-1
= -(x^2+2x+1)^-2 * (2x+2)

Hopefully that's a great enrichment for you in your class. You'll get ahead of the game and see some great applications from simple algebraic changes and a little more knowledge in calculus :)

Ahhh... :)
Got it! Thanks a lot. I'll try to see if i can get this. I'll look up some examples.
Thanks mate

## What is the quotient rule?

The quotient rule is a mathematical rule used to find the derivative of a quotient (or division) of two functions. It states that the derivative of f(x)/g(x) is equal to (g(x)f'(x)-f(x)g'(x))/g(x)^2.

## How do you apply the quotient rule?

To apply the quotient rule, you first identify the two functions in the quotient: f(x) and g(x). Then, you find the derivatives of each function, f'(x) and g'(x). Finally, you plug these values into the quotient rule formula: (g(x)f'(x)-f(x)g'(x))/g(x)^2.

## What is the most common mistake when using the quotient rule?

The most common mistake when using the quotient rule is incorrectly differentiating the numerator and denominator. It is important to remember that the derivative of a quotient is not equal to the quotient of the derivatives.

## Can the quotient rule be used for any type of quotient?

No, the quotient rule can only be used for quotients of two functions. It cannot be used for quotients of multiple functions or for quotients that include constants.

## Are there any alternative methods to find derivatives of quotients?

Yes, there are alternative methods such as the product rule or the chain rule that can also be used to find the derivative of a quotient. However, the quotient rule is specifically designed for quotient functions and is often the most efficient method to use.

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