Consider the following. 1- 2x y = x4 - 2x2 + 4 Let u(x) = 1 - 2x? and v(x) = x* - 2x + 4. Find each indicated derivative. u'(x) = v'(x) Find each indicated product. v(x) · u'(x) u(x) · v'(x) dy Find dx %3D dx

Given:

$\begin{array}{rcl}y& =& \frac{1-2x^2}{x^4-2x^2+4}\\ u\left(x\right)& =& 1-2x^2\\ v\left(x\right)& =& x^4-2x^2+4\end{array}$

We have to find

$$\begin{gathered} 1.u\text{'}\left(x\right) \\ 2.v\text{'}\left(x\right) \\ 3.v\left(x\right)·u\text{'}\left(x\right) \\ 4.u\left(x\right)v\text{'}\left(x\right) \\ 5.\frac{dy}{dx} \end{gathered}$$

The quotient rule of derivative states that $\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left(g\left(x\right)\right)}^2}$ where $f\left(x\right) \text{and} g\left(x\right)$ are differentiable function.

The product rule of derivative states that $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=g\left(x\right)f\text{'}\left(x\right)+f\left(x\right)g\text{'}\left(x\right),$ where $f\left(x\right) \text{and} g\left(x\right)$ are differentiable function.

The sum and difference rule of derivative states that $\frac{d}{dx}\left(f\left(x\right)-g\left(x\right)\right)=f\text{'}\left(x\right)-g\text{'}\left(x\right)$, where $f\left(x\right) \text{and} g\left(x\right)$ are differentiable function.

The power rule of derivative states that $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$.

The constant multiple rule of derivative states that $\frac{d}{dx}\left(cf\left(x\right)\right)=cf\text{'}\left(x\right),$where $f\left(x\right)$ is a differentiable function.

The chain rule of derivative states that $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)$, where $f\left(x\right) \text{and} g\left(x\right)$ are differentiable function.

The derivative of a constant is zero.