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hw11 differentiation rules ii (target c1, c2, c5; §3.3) score: 7/8 answ…

Question

hw11 differentiation rules ii (target c1, c2, c5; §3.3)
score: 7/8 answered: 7/8
question 8
given that
f(x)=x^8h(x)
h(-1)=4
h(-1)=7
calculate f(-1).
hint: use the product rule and the power rule.
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Explanation:

Step1: Apply product rule

The product rule states that if $f(x)=u(x)v(x)$, then $f'(x)=u'(x)v(x)+u(x)v'(x)$. Here $u(x)=x^{8}$ and $v(x)=h(x)$, so $f'(x)=(x^{8})'h(x)+x^{8}h'(x)$.

Step2: Differentiate $x^{8}$

By the power - rule, if $y = x^{n}$, then $y'=nx^{n - 1}$. So, $(x^{8})'=8x^{7}$. Then $f'(x)=8x^{7}h(x)+x^{8}h'(x)$.

Step3: Substitute $x=-1$

Substitute $x = - 1$ into $f'(x)$:
\[

$$\begin{align*} f'(-1)&=8(-1)^{7}h(-1)+(-1)^{8}h'(-1)\\ &=8\times(-1)\times4 + 1\times7\\ &=-32 + 7\\ &=-25 \end{align*}$$

\]

Answer:

$-25$