hw11 differentiation rules ii (target c1, c2, c5; §3.3)\nscore: 7/8 answered: 7/8\nquestion 8\ngiven…

hw11 differentiation rules ii (target c1, c2, c5; §3.3)\nscore: 7/8 answered: 7/8\nquestion 8\ngiven that\nf(x)=x^8h(x)\nh(-1)=4\nh(-1)=7\ncalculate f(-1).\nhint: use the product rule and the power rule.\nquestion help: video message instructor\nsubmit question

hw11 differentiation rules ii (target c1, c2, c5; §3.3)\nscore: 7/8 answered: 7/8\nquestion 8\ngiven that\nf(x)=x^8h(x)\nh(-1)=4\nh(-1)=7\ncalculate f(-1).\nhint: use the product rule and the power rule.\nquestion help: video message instructor\nsubmit question

Answer

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:

$-25$