Question

differentiate.
( y = \frac{8}{1 - 6x^2} )
( \frac{dy}{dx} = square )

Explanation:

Step1: Identify the rule

We use the quotient rule for differentiation. The quotient rule states that if we have a function \( y=\frac{u}{v} \), then \( \frac{dy}{dx}=\frac{u'v - uv'}{v^{2}} \), where \( u = 8 \) and \( v=1 - 6x^{2} \)

Step2: Find \( u' \) and \( v' \)

First, find the derivative of \( u \) with respect to \( x \). Since \( u = 8 \) (a constant), \( u'=\frac{d}{dx}(8) = 0 \)

Next, find the derivative of \( v \) with respect to \( x \). Given \( v = 1-6x^{2} \), using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \), we have \( v'=\frac{d}{dx}(1-6x^{2})=\frac{d}{dx}(1)-6\frac{d}{dx}(x^{2})=0- 12x=-12x \)

Step3: Apply the quotient rule

Substitute \( u = 8 \), \( u'=0 \), \( v = 1-6x^{2} \) and \( v'=- 12x \) into the quotient rule formula:

\( \frac{dy}{dx}=\frac{u'v-uv'}{v^{2}}=\frac{0\times(1 - 6x^{2})-8\times(-12x)}{(1 - 6x^{2})^{2}} \)

Simplify the numerator: \( 0+96x = 96x \)

So, \( \frac{dy}{dx}=\frac{96x}{(1 - 6x^{2})^{2}} \)

Answer:

\( \dfrac{96x}{(1 - 6x^{2})^{2}} \)