Question

question given the function $y = \frac{4x^{2}-1}{2x^{2}-1}$, find $\frac{dy}{dx}$ in simplified form.

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4x^{2}-1$, $u'=8x$, $v = 2x^{2}-1$, and $v' = 4x$.

Step2: Substitute values into quotient - rule

$\frac{dy}{dx}=\frac{(8x)(2x^{2}-1)-(4x^{2}-1)(4x)}{(2x^{2}-1)^{2}}$.

Step3: Expand the numerator

Expand $(8x)(2x^{2}-1)=16x^{3}-8x$ and $(4x^{2}-1)(4x)=16x^{3}-4x$. Then the numerator is $16x^{3}-8x-(16x^{3}-4x)=16x^{3}-8x - 16x^{3}+4x=-4x$.

Answer:

$\frac{-4x}{(2x^{2}-1)^{2}}$