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?
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?
Next time show your work. You probably made a mistake somewhere...
$\displaystyle y = \frac{x^2}{x^2+3}$
$\displaystyle \frac{dy}{dx} = \frac{\left(x^2+3\right)(2x)-x^2(2x)}{\left(x^2+3\right)^2}$
$\displaystyle = \frac{2x^3+6x-2x^3}{\left(x^2+3\right)^2}$
$\displaystyle = \frac{6x}{\left(x^2+3\right)^2}$
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.