§2.4 Derivatives of other trigonometric functions (continued)

Summary

For $f(x) = \sin(x)\cot(x)$, which expression does the product rule give for $f'(x)$?

Using the identity $\sin(x)\cot(x) = \cos(x)$ where sine is nonzero, compute $f'(\frac{\pi}{2})$ for $f(x) = \sin(x)\cot(x)$.

The equation $\sin(x)\cot(x) = \cos(x)$ holds for every real number $x$.

An object has height $h(t) = 3 + \frac{2\cos(t)}{1.2^t}$ feet at time $t$ seconds, with $h'(2) \approx -1.16$ and $h''(2) \approx 1.02$. At $t = 2$ the object is

For $h(t) = 3 + \frac{2\cos(t)}{1.2^t}$, compute the velocity $h'(2)$ rounded to four decimals.

For $p(z) = \frac{z\tan(z)}{z^2\sec(z) + 1} + 3e^z + 1$, the tangent line at $z = 0$ is

For the same $p$, since $p'(0) = 3 \gt 0$, the function is increasing at $z = 0$.

Differentiating $h(t) = 3 + \frac{2\cos(t)}{1.2^t}$ twice gives $h''(2) \approx 1.0193$ feet per second per second.