17. -/1 points details my notes larcalc12m 2.3.016.mi. use the quotient rule to find the derivative of the…

17.

Answer:

$-\frac{t\sin(t)+ 7\cos(t)}{t^{8}}$

Explanation:

Step1: Recall quotient - rule

The quotient rule states that if $y = \frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u=\cos(t)$ and $v = t^{7}$.

Step2: Find $u'$ and $v'$

The derivative of $u=\cos(t)$ is $u'=-\sin(t)$, and the derivative of $v = t^{7}$ using the power - rule $(x^{n})'=nx^{n - 1}$ is $v' = 7t^{6}$.

Step3: Apply quotient - rule

[ \begin{align*} f'(t)&=\frac{(-\sin(t))\cdot t^{7}-\cos(t)\cdot7t^{6}}{(t^{7})^{2}}\ &=\frac{-t^{7}\sin(t)-7t^{6}\cos(t)}{t^{14}}\ &=\frac{t^{6}(-t\sin(t)-7\cos(t))}{t^{14}}\ &=-\frac{t\sin(t)+ 7\cos(t)}{t^{8}} \end{align*} ]

18.

Answer:

$2x + 15x^{2}$

Explanation:

Step1: Recall sum - rule of differentiation

If $g(x)=u(x)+v(x)$, then $g'(x)=u'(x)+v'(x)$. Here, $u(x)=x^{2}$ and $v(x)=5x^{3}$.

Step2: Apply power - rule

The power - rule states that $(x^{n})'=nx^{n - 1}$. For $u(x)=x^{2}$, $u'(x)=2x$, and for $v(x)=5x^{3}$, $v'(x)=5\times3x^{2}=15x^{2}$.

Step3: Find $g'(x)$

$g'(x)=u'(x)+v'(x)=2x + 15x^{2}$

19.

Answer:

$-\frac{15s^{2}}{(s^{3}-4)^{4}}$

Explanation:

Step1: Rewrite the function

We can rewrite $g(s)=\frac{5}{(s^{3}-4)^{3}}=5(s^{3}-4)^{-3}$.

Step2: Apply chain - rule

The chain - rule states that if $y = f(u)$ and $u = h(x)$, then $y'=f'(u)\cdot h'(x)$. Let $u = s^{3}-4$, so $y = 5u^{-3}$. The derivative of $y$ with respect to $u$ is $y_{u}'=-15u^{-4}$, and the derivative of $u$ with respect to $s$ is $u_{s}' = 3s^{2}$.

Step3: Calculate $g'(s)$

[ \begin{align*} g'(s)&=y_{u}'\cdot u_{s}'\ &=-15(s^{3}-4)^{-4}\cdot3s^{2}\ &=-\frac{15s^{2}}{(s^{3}-4)^{4}} \end{align*} ]