# Derivative of $\frac{8x^2}{3\left(x^2+1\right)^3}$ when it equals $0$.

I want to calculate the derivative of

$$\frac{8x^2}{3\left(x^2+1\right)^3}$$

And I get by equation

$$\left[\frac{f(x)}{g(x)}\right]' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$

that the derivative is

$$\frac{3\cdot16x\left(x^2+1\right)^3 - 8x^2\cdot18x\left(x^2+1\right)^2}{9\left(x^2+1\right)^6}$$

By simplification it should be equal to:

$$\frac{3\cdot16x(x^2+1) - 8x^2\cdot18x}{9\left(x^2+1\right)^4}$$

Then I want to equal the derivative to $0$.

$$\frac{48x(x^2 + 1) - 144x^3}{9\left(x^2+1\right)^4} = 0$$

$$48x^3 + 48x - 144x^3 = 0$$

$$-36x^3 + 48x = 0$$

$$x(-36x^2 + 48) = 0$$

The solutions would be

x = 0,\qquad \begin{align}&48 = 36x^2 \\ &x^2 = \frac43 \end{align}

This finishes in a wrong answer.

The solution I have with me resolve it without simplification. Nevertheless, it should come a correct answer also with my way. So what's wrong. The solution I have is:

$$f'(x) = \frac{16x\cdot3\left(x^2 + 1\right)^3 - 8x^2\cdot9\left(x^2+1\right)^2\cdot2x}{\left(3\left(x^2+1\right)^3\right)^2}$$

Now we will equal it to $0$.

$$48x\left(x^2 + 1\right)^3 - 144x^3\left(x^2+1\right)^2 = 0$$ $$48x\left(x^2 + 1\right)^2\left[x^2 + 1 - 3x^2\right] = 0$$

That means that $x = 0$ or $$-2x^2 + 1 = 0$$ That means that $$x = \frac{1}{\sqrt{2}}$$

(It is given that $x$ is positive).

The error was in your first attempts where you simplified $48x^3+48x-144x^3=0$ to $48x-36x^3=0$. It should be $48x-96x^3=0$