§2.1 Elementary derivative rules (continued)

Summary

Differentiable functions $f$ and $g$ satisfy $f(2) = 5$, $g(2) = -3$, $f'(2) = -\frac{1}{2}$, and $g'(2) = 2$. For $h(x) = 3f(x) - 4g(x)$, what is $h'(2)$?

With the same functions $f$ and $g$ where $f(2) = 5$ and $g(2) = -3$, compute $h(2)$ for $h(x) = 3f(x) - 4g(x)$.

The tangent line to $h(x) = 3f(x) - 4g(x)$ at $x = 2$ is $L(x) = 27 - \frac{19}{2}(x - 2)$. Use it to estimate $h(2.1)$, rounded to two decimals.

With the same data, let $p(x) = -2f(x) + \frac{1}{2}g(x)$. Since $p'(2) = -2f'(2) + \frac{1}{2}g'(2) = 1 + 1 = 2$, the function $p$ at $a = 2$ is

With $p(2) = -2(5) + \frac{1}{2}(-3) = -11.5$ and $p'(2) = 2$, estimate $p(2.03)$ using the local linearization, rounded to two decimals.

Piecewise linear functions $p$ and $q$ both have sharp corners at $x = -1$ and $x = 1$ and are smooth elsewhere. For $r(x) = p(x) + 2q(x)$, where is $r$ not differentiable?

For the same $r(x) = p(x) + 2q(x)$, the graphs show $p'(-2) = -2$ and $q'(-2) = 3$. Compute $r'(-2)$.

For the same $r$, the graphs give $r(2) = 4$ and $r'(2) = p'(2) + 2q'(2) = 0$. The tangent line to $y = r(x)$ at $(2, r(2))$ is

Let $r(t) = t^t$ and $s(t) = \arccos(t)$ with given formulas $r'(t) = t^t(\ln t + 1)$ and $s'(t) = -\frac{1}{\sqrt{1-t^2}}$. For $w(t) = 3t^t - 2\arccos(t)$, which expression is $w'(t)$?

With $v(t) = t^t + \arccos(t)$ and $v'(\frac{1}{2}) \approx -0.938$, the function $v$ is decreasing at the instant $t = \frac{1}{2}$.

Starting from the limit definition, the derivative of $f(x) = a^x$ (with $a \gt 0$) turns out to be

Numerical experiments give $\lim_{h \to 0}\frac{2^h - 1}{h} \approx 0.6931$ and $\lim_{h \to 0}\frac{3^h - 1}{h} \approx 1.0986$. For which base does the limit equal exactly $1$, making the derivative of $a^x$ equal $a^x$?