Question

differentiate: v(x)=(2x^3 - 6)(x^4 - 6x) v(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $V(x)=u(x)v(x)$, then $V^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Let $u(x)=2x^{3}-6$ and $v(x)=x^{4}-6x$.

Step2: Differentiate $u(x)$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u^{\prime}(x)=\frac{d}{dx}(2x^{3}-6)=6x^{2}$.

Step3: Differentiate $v(x)$

Using the power - rule, $v^{\prime}(x)=\frac{d}{dx}(x^{4}-6x)=4x^{3}-6$.

Step4: Substitute into product - rule

$V^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)=6x^{2}(x^{4}-6x)+(2x^{3}-6)(4x^{3}-6)$.

Step5: Expand the expressions

First expand $6x^{2}(x^{4}-6x)=6x^{6}-36x^{3}$. Then expand $(2x^{3}-6)(4x^{3}-6)=8x^{6}-12x^{3}-24x^{3}+36 = 8x^{6}-36x^{3}+36$.

Step6: Combine like terms

$V^{\prime}(x)=(6x^{6}-36x^{3})+(8x^{6}-36x^{3}+36)=14x^{6}-72x^{3}+36$.

Answer:

$14x^{6}-72x^{3}+36$