Question

question: for ( j(x)=(6 - x^{2})(-8x^{3}-7x)), find ( j(x)) by apply provide your answer below: (j(x)=square)

Explanation:

Step1: Recall product - rule

The product - rule states that if \(y = u(x)v(x)\), then \(y'=u'(x)v(x)+u(x)v'(x)\). Let \(u(x)=6 - x^{2}\) and \(v(x)=-8x^{3}-7x\).

Step2: Find \(u'(x)\)

Differentiate \(u(x)=6 - x^{2}\) with respect to \(x\). Using the power - rule \((x^{n})'=nx^{n - 1}\) and the constant - rule \((c)' = 0\) (\(c\) is a constant), we get \(u'(x)=0-2x=-2x\).

Step3: Find \(v'(x)\)

Differentiate \(v(x)=-8x^{3}-7x\) with respect to \(x\). Using the power - rule \((x^{n})'=nx^{n - 1}\), we have \(v'(x)=-8\times3x^{2}-7=-24x^{2}-7\).

Step4: Apply the product - rule

\[

$$\begin{align*} j'(x)&=u'(x)v(x)+u(x)v'(x)\\ &=(-2x)(-8x^{3}-7x)+(6 - x^{2})(-24x^{2}-7)\\ &=(-2x)\times(-8x^{3})+(-2x)\times(-7x)+6\times(-24x^{2})+6\times(-7)-x^{2}\times(-24x^{2})-x^{2}\times(-7)\\ &=16x^{4}+14x^{2}-144x^{2}-42 + 24x^{4}+7x^{2}\\ &=(16x^{4}+24x^{4})+(14x^{2}-144x^{2}+7x^{2})-42\\ &=40x^{4}-123x^{2}-42 \end{align*}$$

\]

Answer:

\(40x^{4}-123x^{2}-42\)