4. given: $f(x)=(2x^{6}+3)(4x^{2}-1 - 5x)$\n a) apply the product rule to find the derivative. fill in the…

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 $u(x)=2x^{6}+3$ and $v(x)=4x^{2}-1 - 5x$. Then $u'(x)=12x^{5}$ and $v'(x)=8x - 5$. So $f'(x)=(12x^{5})(4x^{2}-1 - 5x)+(2x^{6}+3)(8x - 5)$.

Step2: Multiply factors of $f(x)$

[ \begin{align*} f(x)&=(2x^{6}+3)(4x^{2}-1 - 5x)\ &=2x^{6}(4x^{2}-1 - 5x)+3(4x^{2}-1 - 5x)\ &=8x^{8}-2x^{6}-10x^{7}+12x^{2}-3 - 15x\ &=8x^{8}-10x^{7}-2x^{6}+12x^{2}-15x - 3 \end{align*} ] Then, using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=64x^{7}-70x^{6}-12x^{5}+24x-15$.

Step3: Evaluate $f'(1)$ for part a

[ \begin{align*} f'(1)&=(12\times1^{5})(4\times1^{2}-1 - 5\times1)+(2\times1^{6}+3)(8\times1 - 5)\ &=12\times(4 - 1 - 5)+(2 + 3)\times(8 - 5)\ &=12\times(-2)+5\times3\ &=-24 + 15\ &=-9 \end{align*} ]

Step4: Evaluate $f'(1)$ for part b

[ \begin{align*} f'(1)&=64\times1^{7}-70\times1^{6}-12\times1^{5}+24\times1-15\ &=64-70-12 + 24-15\ &=(64 + 24)-(70+12 + 15)\ &=88 - 97\ &=-9 \end{align*} ] We notice that the values of $f'(1)$ obtained from parts a and b are the same. This happens because the product rule and the method of first expanding the function and then differentiating are two equivalent ways of finding the derivative of a function.

Answer:

a) $f'(x)=(12x^{5})(4x^{2}-1 - 5x)+(2x^{6}+3)(8x - 5)$ b) $f(x)=8x^{8}-10x^{7}-2x^{6}+12x^{2}-15x - 3$, $f'(x)=64x^{7}-70x^{6}-12x^{5}+24x-15$ c) $f'(1)=-9$ (from both part a and part b). The values are the same because the two methods of finding the derivative are equivalent.