Corrected exercises: derivatives of trigonometric functions (MATH 203)

Sum and constant multiple rules, term by term: $f'(x) = 7\cos x - 4(-\csc x\cot x) + 6x = 7\cos x + 4\csc x\cot x + 6x$. The minus written in front of $\csc x$ meets the minus of $(\csc x)'$, and the two make a plus: this is the sign slip of the chapter. At $\frac{\pi}{6}$: $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$, $\csc\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = 2$, $\cot\frac{\pi}{6} = \frac{\cos(\pi/6)}{\sin(\pi/6)} = \sqrt 3$. So $f'\left(\frac{\pi}{6}\right) = \frac{7\sqrt 3}{2} + 4 \cdot 2\sqrt 3 + 6\cdot\frac{\pi}{6} = \frac{7\sqrt 3}{2} + \frac{16\sqrt 3}{2} + \pi = \frac{23\sqrt 3}{2} + \pi$, about $23.0602$.

Rewrite the numerator in cosines and bring it to one fraction: $\sec x - \cos x = \frac{1}{\cos x} - \frac{\cos^2 x}{\cos x} = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$, by the Pythagorean identity. Dividing by $\sin x$ is multiplying by $\frac{1}{\sin x}$: $g(x) = \frac{\sin^2 x}{\cos x\sin x} = \frac{\sin x}{\cos x} = \tan x$, wherever $\sin x \ne 0$ and $\cos x \ne 0$. Hence $g'(x) = \sec^2 x$ and $g'\left(\frac{\pi}{3}\right) = \frac{1}{\cos^2(\pi/3)} = \frac{1}{1/4} = 4$. The quotient rule applied to the original form also works, but it takes half a page and three chances to lose a sign. The formula $\sec^2 x$ is valid only on the domain of $g$: at $x = \pi$, where $g$ does not exist, neither does $g'$.

Use the disguise $\tan^2 x = \sec^2 x - 1$: the numerator becomes $\sec^2 x - 1 = (\sec x - 1)(\sec x + 1)$, a difference of squares. The factor $\sec x + 1$ cancels with the denominator, which is legitimate wherever it is not $0$, that is $\cos x \ne -1$; and $\cos x \ne 0$ for $\tan x$ and $\sec x$ to exist. There $h(x) = \sec x - 1$, so $h'(x) = \sec x\tan x$ and $h'\left(\frac{\pi}{3}\right) = 2\cdot\sqrt 3 = 2\sqrt 3$, about $3.4641$. The gesture is to see $\tan^2 x$ as $\sec^2 x - 1$ BECAUSE the denominator is written with $\sec x$: the identity is chosen by the rest of the fraction. The quotient rule on the original form works too, but it needs $(\tan^2 x)'$, a product, and a page of simplification.

Term by term: $k'(x) = \sec^2 x - \csc^2 x = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}$. Common denominator $\sin^2 x\cos^2 x$: $k'(x) = \frac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x}$. At $\frac{\pi}{4}$, $\sin^2 x = \cos^2 x = \frac{1}{2}$, so $k'\left(\frac{\pi}{4}\right) = 0$: the graph of $k$ has a horizontal tangent there. At $\frac{\pi}{6}$, $\sin^2 x = \frac{1}{4}$ and $\cos^2 x = \frac{3}{4}$: $k'\left(\frac{\pi}{6}\right) = \frac{1/4 - 3/4}{3/16} = -\frac{1}{2}\cdot\frac{16}{3} = -\frac{8}{3}$. Check with the first form: $\sec^2\frac{\pi}{6} = \frac{4}{3}$ and $\csc^2\frac{\pi}{6} = 4$, and $\frac{4}{3} - 4 = -\frac{8}{3}$. The minus of $(\cot x)'$ is what produces a difference and not a sum.

Common denominator $(1 + \sin x)\cos x$: the numerator is $\cos^2 x + (1 + \sin x)^2 = \cos^2 x + 1 + 2\sin x + \sin^2 x$. Collect $\sin^2 x + \cos^2 x = 1$: the numerator is $2 + 2\sin x = 2(1 + \sin x)$. Cancel the FACTOR $1 + \sin x$, which is not zero where $m$ is defined (if $\sin x = -1$ then $\cos x = 0$): $m(x) = \frac{2(1 + \sin x)}{(1 + \sin x)\cos x} = \frac{2}{\cos x} = 2\sec x$. Hence $m'(x) = 2\sec x\tan x$ and $m'\left(\frac{\pi}{4}\right) = 2\cdot\sqrt 2\cdot 1 = 2\sqrt 2$, about $2.8284$. Three algebra gestures in a row, expand the square, spot the identity, factor the $2$: this is the exact place where a MATH 203 derivative is won or lost.

$u = x^3$, $u' = 3x^2$, $v = \cos x$, $v' = -\sin x$. So $y' = 3x^2\cos x + x^3(-\sin x) = 3x^2\cos x - x^3\sin x = x^2(3\cos x - x\sin x)$, the smallest power $x^2$ factored out. At $\pi$: $\cos\pi = -1$ and $\sin\pi = 0$, so $y'(\pi) = \pi^2(-3 - 0) = -3\pi^2$, about $-29.6088$. Writing $(x^3\cos x)' = 3x^2(-\sin x)$, the product of the derivatives, is worth nothing.

Rewrite the root as a power first: $\sqrt x = x^{1/2}$, so $u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}$. With $v = \tan x$ and $v' = \sec^2 x$: $y' = \frac{\tan x}{2\sqrt x} + \sqrt x\sec^2 x$. Single fraction over $2\sqrt x$: $\sqrt x\sec^2 x = \frac{2x\sec^2 x}{2\sqrt x}$, because $\sqrt x\cdot 2\sqrt x = 2x$. So $y' = \frac{\tan x + 2x\sec^2 x}{2\sqrt x}$, for $x > 0$ with $\cos x \ne 0$. At $\frac{\pi}{4}$: $\tan = 1$, $\sec^2 = 2$, $2\sqrt{\pi/4} = \sqrt\pi$, so $y'\left(\frac{\pi}{4}\right) = \frac{1 + \pi}{\sqrt\pi}$, about $2.3366$. The negative exponent $x^{-1/2}$ becomes a root in the DENOMINATOR, not $-\sqrt x$.

First product, $u = x^2 - 2$, $v = \sin x$: $2x\sin x + (x^2 - 2)\cos x$. Second product, $u = 2x$, $v = \cos x$: $2\cos x - 2x\sin x$. Add and collect: the terms $2x\sin x$ and $-2x\sin x$ cancel, and $(x^2 - 2)\cos x + 2\cos x = x^2\cos x$. So $y' = x^2\cos x$. At $\frac{\pi}{3}$: $y' = \frac{\pi^2}{9}\cdot\frac{1}{2} = \frac{\pi^2}{18}$, about $0.5483$. A student who stops at the four-term sum has not answered a question that says simplify, and misses the whole point: this function was built to have the derivative $x^2\cos x$.

$u = x$, $v = e^x$, $w = \sin x$, with $u' = 1$, $v' = e^x$, $w' = \cos x$: $y' = e^x\sin x + xe^x\sin x + xe^x\cos x$. Factor $e^x$, then collect the sines: $y' = e^x\left[(1 + x)\sin x + x\cos x\right]$. At $\pi$: $\sin\pi = 0$ and $\cos\pi = -1$, so $y'(\pi) = e^\pi\left[0 - \pi\right] = -\pi e^\pi$, about $-72.6986$. Three factors give three terms, each with exactly ONE derivative in it.

$u = \csc x$, $u' = -\csc x\cot x$, $v = \cot x$, $v' = -\csc^2 x$. So $y' = -\csc x\cot x\cdot\cot x + \csc x\cdot(-\csc^2 x) = -\csc x\cot^2 x - \csc^3 x = -\csc x\left(\cot^2 x + \csc^2 x\right)$. With $\cot^2 x = \csc^2 x - 1$, obtained by dividing $\sin^2 x + \cos^2 x = 1$ by $\sin^2 x$: $y' = -\csc x\left(2\csc^2 x - 1\right)$. At $\frac{\pi}{6}$, $\csc = 2$: $y' = -2(8 - 1) = -14$. Check with the unsimplified form: $\cot^2\frac{\pi}{6} = 3$, and $-2(3 + 4) = -14$. Both derivatives are negative, so both minus signs survive the product rule.

$u = \sin x + \cos x$, $u' = \cos x - \sin x$; $v = \sin x - \cos x$, $v' = \cos x + \sin x$. Numerator: $u'v - uv' = (\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)^2 = -(\sin x - \cos x)^2 - (\sin x + \cos x)^2$. Expand both squares: $(\sin^2 x - 2\sin x\cos x + \cos^2 x) + (\sin^2 x + 2\sin x\cos x + \cos^2 x) = 2(\sin^2 x + \cos^2 x) = 2$. So the numerator is $-2$ and $q'(x) = \frac{-2}{(\sin x - \cos x)^2}$. Then $q'(0) = \frac{-2}{(0 - 1)^2} = -2$. At $\frac{\pi}{4}$, $\sin x = \cos x$: the denominator of $q$ vanishes, $q$ is not defined there and neither is $q'$. The cross terms $\pm 2\sin x\cos x$ cancel: expanding is what makes the identity visible.

$u = 1 - \cos x$, $u' = \sin x$; $v = 1 + \cos x$, $v' = -\sin x$. Numerator: $\sin x(1 + \cos x) - (1 - \cos x)(-\sin x) = \sin x + \sin x\cos x + \sin x - \sin x\cos x = 2\sin x$. So $r'(x) = \frac{2\sin x}{(1 + \cos x)^2}$, for $\cos x \ne -1$. At $\frac{2\pi}{3}$: $\sin = \frac{\sqrt 3}{2}$, $1 + \cos = \frac{1}{2}$, so $r' = \frac{\sqrt 3}{1/4} = 4\sqrt 3$, about $6.9282$. The minus of the rule meets the minus of $v'$: $-(1 - \cos x)(-\sin x)$ is a PLUS. Lose it and the $\sin x\cos x$ terms add instead of cancelling.

$u = \tan x$, $u' = \sec^2 x$, $v = x$, $v' = 1$: $g'(x) = \frac{x\sec^2 x - \tan x}{x^2}$, for $x \ne 0$ and $\cos x \ne 0$. At $\frac{\pi}{4}$: $\frac{\frac{\pi}{4}\cdot 2 - 1}{\pi^2/16} = 16\cdot\frac{\pi/2 - 1}{\pi^2} = \frac{8\pi - 16}{\pi^2}$, about $0.9254$. The order matters: $\frac{\tan x - x\sec^2 x}{x^2}$ gives $-0.9254$, a slope of the wrong sign.

$u = \sec x$, $u' = \sec x\tan x$; $v = \tan x$, $v' = \sec^2 x$. Numerator: $\sec x\tan x\cdot\tan x - \sec x\cdot\sec^2 x = \sec x\left(\tan^2 x - \sec^2 x\right) = -\sec x$, by $\sec^2 x = 1 + \tan^2 x$. So $h'(x) = \frac{-\sec x}{\tan^2 x} = -\frac{1}{\cos x}\cdot\frac{\cos^2 x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x\cot x$. Second road: $h(x) = \frac{1}{\cos x}\cdot\frac{\cos x}{\sin x} = \frac{1}{\sin x} = \csc x$, so $h'(x) = -\csc x\cot x$ at once. But $h$ is NOT defined where $\cos x = 0$, although $\csc x$ is: the formula holds only where $\sin x \ne 0$ and $\cos x \ne 0$. At $\frac{\pi}{3}$: $-\frac{2}{\sqrt 3}\cdot\frac{1}{\sqrt 3} = -\frac{2}{3}$.

Quotient rule, $u = x^2$, $v = \cos x$: $s'(x) = \frac{2x\cos x - x^2(-\sin x)}{\cos^2 x} = \frac{2x\cos x + x^2\sin x}{\cos^2 x}$. Product rule on $x^2\sec x$: $s'(x) = 2x\sec x + x^2\sec x\tan x$. They agree: split the fraction, $\frac{2x\cos x}{\cos^2 x} = \frac{2x}{\cos x} = 2x\sec x$ and $\frac{x^2\sin x}{\cos^2 x} = x^2\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = x^2\sec x\tan x$. At $\frac{\pi}{3}$, $\sec = 2$, $\tan = \sqrt 3$: $s' = 2\cdot\frac{\pi}{3}\cdot 2 + \frac{\pi^2}{9}\cdot 2\sqrt 3 = \frac{4\pi}{3} + \frac{2\sqrt 3\,\pi^2}{9}$, about $7.9876$. When the denominator is a single $\cos x$, the product with $\sec x$ is usually shorter.

Point: $\csc\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = 2$. Slope: $(\csc x)' = -\csc x\cot x$, and $\cot\frac{\pi}{6} = \sqrt 3$, so $m = -2\sqrt 3$. Tangent: $y = 2 - 2\sqrt 3\left(x - \frac{\pi}{6}\right) = -2\sqrt 3\,x + 2 + \frac{2\sqrt 3\,\pi}{6} = -2\sqrt 3\,x + 2 + \frac{\sqrt 3\,\pi}{3}$. The intercept is about $3.8138$. Expanding $-2\sqrt 3\cdot\left(-\frac{\pi}{6}\right)$ is where the sign goes: minus times minus is plus.

$y(\frac{\pi}{2}) = \frac{\pi}{2}\sin\frac{\pi}{2} = \frac{\pi}{2}$. Product rule: $y' = \sin x + x\cos x$, so $y'\left(\frac{\pi}{2}\right) = 1 + \frac{\pi}{2}\cdot 0 = 1$. Tangent: $y = \frac{\pi}{2} + \left(x - \frac{\pi}{2}\right) = x$. Normal: slope $-\frac{1}{1} = -1$, so $y = \frac{\pi}{2} - \left(x - \frac{\pi}{2}\right) = \pi - x$. It meets the $x$-axis at $(\pi, 0)$, and $\pi\sin\pi = 0$: that point is on the curve. The solution figure is drawn with equal scales, so the right angle between tangent and normal is true.