1. ## Quotient Rule Help

Hey.

I was not able to make the class where the explained the quotient rule in calculus.

I'm trying to self learn it but the book isn't helping that much.

They're trying to find the derivative of:

$y\prime=\frac{x^2+x-2}{x^3+6}$

They use the Quotient Rule Formula to get to

$y\prime=\frac{{(x^3+6)}{\frac{d}{dx}}{(x^2+x-2)}-{(x^2+x-2)}{\frac{d}{dx}}{(x^3+6)}}{{(x^3+6)}^2}$

Thats all well and good, but then they make the jump from that to:

$y\prime=\frac{(x^3+6)(2x+1)-(x^2+x-2)(3x^2)}{(x^3+6)^2}$

And then they just simplify that to:

$y\prime=\frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}$

But can someone explain that first jump?

Thanks!

2. y'= (x^3+6) * (d/dx)(x^2+x-2) - (x^2+x-2) *(d/dx)(x^3+6)/(x^3+6)^2

You take the derivative of the one behind the d/dx so you get:
(x^3+6)(2x+1) - (x^2+x-2)(3x^2)/(x^2+6)^2

so in other words you take the derivative of (x^2+x-2) to get (2x+1) and you take the derivative of (x^3+6) to get (3x^2)

I hope this helps

3. Oh so the d/dx means to take the derivative of?

4. Originally Posted by Saphric
Oh so the d/dx means to take the derivative of?
Yes. Have you not met that notation before?

5. Originally Posted by Saphric
Hey.

I was not able to make the class where the explained the quotient rule in calculus.

I'm trying to self learn it but the book isn't helping that much.

They're trying to find the derivative of:

$y\prime=\frac{x^2+x-2}{x^3+6}$

I am not a fan of the quotient rule - too much confusion, so convert it to a product...

$y=(x^2+x-2)(x^3+6)^{-1}$

$y'=(2x+1)(x^3+6)^{-1} - (x^2+x-2)(3x^2)(x^3+6)^{-2}$

$y'=\frac{(2x+1)(x^3+6)-(x^2+x-2)(3x^2)}{(x^3+6)^2}$

$y'=\frac{2x^4+12x+x^3+6-3x^4-3x^3-6x^2}{(x^3+6)^2} \Rightarrow \frac{-x^4-2x^3-6x^2+12x+6}{(x^3+6)^2}$