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)