§2.3 The product and quotient rules (continued)

Combining rules

Summary

To differentiate $f(x) = x\sin(x) + \frac{x^2}{\cos(x) + 2}$, which combination of rules is required?

To differentiate $s(y) = \frac{y \cdot 7^y}{y^2 + 1}$, which combination of rules is required?

Let $f(r) = (5r^3 + \sin(r))(4^r - 2\cos(r))$. Which structure does $f'(r)$ have?

Let $p(t) = \frac{\cos(t)}{t^6 \cdot 6^t}$. Which structure does $p'(t)$ have?

For $g(z) = 3z^7e^z - 2z^2\sin(z) + \frac{z}{z^2+1}$, compute $g'(0)$.

Differentiable $f$ and $g$ satisfy $f(3) = -2$, $f'(3) = 7$, $g(3) = 4$, and $g'(3) = -1$. For $p(x) = f(x)g(x)$, compute $p'(3)$.

With the same data $f(3) = -2$, $f'(3) = 7$, $g(3) = 4$, $g'(3) = -1$, and $q(x) = \frac{f(x)}{g(x)}$, compute $q'(3)$.

Suppose $f(2) = 5$, $g(2) = -3$, $f'(2) = -\frac{1}{2}$, and $g'(2) = 2$. For $h(x) = g(x)f(x)$, compute $h'(2)$ as a decimal.

With the same data, let $r(x) = \frac{g(x)}{f(x)}$, so $r(2) = -\frac{3}{5}$ and $r'(2) = \frac{17}{50}$. Use the local linearization to estimate $r(2.06)$, rounded to four decimals.

Piecewise linear functions $p$ and $q$ have corners at $x = -1$ and $x = 1$ and are differentiable elsewhere. For $r(x) = p(x)q(x)$ and $z(x) = \frac{q(x)}{p(x)}$, with $p(1) = 0$, where do $r'$ and $z'$ fail to exist?

A farmer plants $A(t)$ thousand acres with yield $Y(t)$ hundred bushels per thousand acres, so the harvest is $C(t) = A(t)Y(t)$. The rate $C'(t)$ equals

The farmer has $A(0) = 7000$ acres, $Y(0) = 170$ bushels per acre, $A'(0) = 600$ acres per year, and $Y'(0) = 8$ bushels per acre per year. Compute $C'(0)$ in bushels per year.

A car uses $f(v)$ liters per kilometer at speed $v$, with $f(80) = 0.05$ and $f'(80) = 0.0004$. The distance per liter $g(v) = \frac{1}{f(v)}$ satisfies

The same car's hourly consumption is $h(v) = v \cdot f(v)$ liters per hour. With $f(80) = 0.05$ and $f'(80) = 0.0004$, the values $h(80)$ and $h'(80)$ are

The product rule formula for $f(x)g(x)$ is valid at a point only if both $f$ and $g$ are differentiable at that point.

With $r(t) = t^t$ and $s(t) = \arccos(t)$ and given derivative formulas, the derivatives of $w(t) = t^t\arccos(t)$ and $v(t) = \frac{t^t}{\arccos(t)}$ follow from the product and quotient rules respectively.