Question

given the function $y = \frac{1 - x}{2x^{3}-5}$, find $\frac{dy}{dx}$ in simplified form.

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1 - x$, so $u^\prime=-1$; $v = 2x^{3}-5$, so $v^\prime = 6x^{2}$.

Step2: Substitute into quotient - rule formula

$y^\prime=\frac{(-1)\times(2x^{3}-5)-(1 - x)\times(6x^{2})}{(2x^{3}-5)^{2}}$.

Step3: Expand the numerator

\[

$$\begin{align*} (-1)\times(2x^{3}-5)-(1 - x)\times(6x^{2})&=-2x^{3}+5-(6x^{2}-6x^{3})\\ &=-2x^{3}+5 - 6x^{2}+6x^{3}\\ &=4x^{3}-6x^{2}+5 \end{align*}$$

\]

Answer:

$\frac{4x^{3}-6x^{2}+5}{(2x^{3}-5)^{2}}$