I need help finding the derivative of this. I am a bit confused. I have y = (x^2)/(x^2+3). the derivative is 6x/(x^2+3)^2. I understand the quotient rule, and how they got the squared on the bottom. but how did they get the 6x on the top?

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- Jun 13th 2012, 11:24 AMdroidusfinding derivative
I need help finding the derivative of this. I am a bit confused. I have y = (x^2)/(x^2+3). the derivative is 6x/(x^2+3)^2. I understand the quotient rule, and how they got the squared on the bottom. but how did they get the 6x on the top?

- Jun 13th 2012, 11:51 AMReckonerRe: finding derivative
Next time show your work. You probably made a mistake somewhere...

$\displaystyle y = \frac{x^2}{x^2+3}$

$\displaystyle \frac{dy}{dx} = \frac{\left(x^2+3\right)(2x)-x^2(2x)}{\left(x^2+3\right)^2}$

$\displaystyle = \frac{2x^3+6x-2x^3}{\left(x^2+3\right)^2}$

$\displaystyle = \frac{6x}{\left(x^2+3\right)^2}$ - Jun 13th 2012, 01:14 PMdroidusRe: finding derivative
ok - here is what i get for the second derivative then...

[(x^2 + 3)^2 + 6 - 6x(4x)] / (x^2+3)^4 - Jun 13th 2012, 01:29 PMReckonerRe: finding derivative
The derivative of $\displaystyle \left(x^2+3\right)^2$ is not $\displaystyle 4x.$ And shouldn't you be multiplying that 6 instead of adding it? Let's be more careful:

$\displaystyle \frac{dy}{dx} = \frac{6x}{\left(x^2+3\right)^2}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{\left(x^2+3\right)^2\cdot\frac d{dx}[6x] - 6x\cdot\frac d{dx}\left[\left(x^2+3\right)^2\right]}{\left(x^2+3\right)^4}$

$\displaystyle = \frac{6\left(x^2+3\right)^2 - 6x\cdot2\left(x^2+3\right)\cdot\frac d{dx}\left[x^2+3\right]}{\left(x^2+3\right)^4}$

$\displaystyle = \frac{6\left(x^2+3\right)^2 - 24x^2\left(x^2+3\right)}{\left(x^2+3\right)^4}$

$\displaystyle = \frac{6\left(x^2+3\right) - 24x^2}{\left(x^2+3\right)^3}$

$\displaystyle = \frac{-18\left(x^2-1\right)}{\left(x^2+3\right)^3}$

$\displaystyle = \frac{-18(x-1)(x+1)}{\left(x^2+3\right)^3}$

Make sure that you're using the chain rule properly. We must remember to differentiate the "inner" function after we differentiate the "outer" function of a composite function.