Derivatives of a fraction function

Question

An example of a fraction function is: $$y= \frac{-8x}{(x^2 + 3)^2}$$ The quotient rule says that if the function one wishes to differentiate, $f(x)$, can be written as: $$h(x) = \frac{f(x)}{g(x)}$$ Then the derivative is (according to what I learned): $$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$

Then I think the procedure is the following:

\begin{align} y' & = \frac{24(x^2 - 1)}{((x^2+3)^2)^2}\\ & = \frac{24(x^2 - 1)}{(x^2+3)^4}\\ \end{align}

However, the solution is... $$y'= \frac{24(x^2 - 1)}{(x^2+3)^3}$$

• What are my mistakes?
• What is the correct way to derivate fractions?

Answer 1

Your second step (after writing down the quotient rule) should be: $$ y' = \frac{-8(x^2+3)^2+8x \cdot 2 \cdot 2x(x^2+3)}{((x^2+3)^2)^2}, $$ and then an $x^2+3$ cancels off and gives you the correct answer: $$ \frac{-8(x^2+3)^2+8x \cdot 2 \cdot 2x(x^2+3)}{((x^2+3)^2)^2} = \frac{-8(x^2+3)+32x^2}{(x^2+3)^3} = \frac{24(x^2-1)}{(x^2+3)^3}. $$

Answer 2

Hint #1: If you write your fraction as $$HIGH/LOW,$$ then the derivative for the quotient is "given" by the mnemonic $$\frac{LOW\cdot{d(HIGH)} - HIGH \cdot{d(LOW)}}{(LOW)^2}.$$

Hint #2: $d(HIGH) = -8$

Hint #3: $d(LOW) = 2(x^2 + 3)(2x) = 4x(x^2 + 3)$

Can you take it from here?