Derive Quotient Rule: Find the Mistake

[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

 

[silvermane]

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

 

[A_Munk3y]

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

 

[silvermane]

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 :)

 

[Quinzio]

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 ??

 

[silvermane]

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?

 

[Quinzio]

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

 

[A_Munk3y]

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

 

[silvermane]

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 :)

 

[A_Munk3y]

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