- Which squared reciprocal-trigonometric functions appear in these derivative rules? - Check the sign of each formula rather than assuming they match.

1. The derivative of \(\tan x\) is \(\sec^2 x\). 2. The derivative of \(\cot x\) is \(-\csc^2 x\).

a) \(f'(x)=\sec^2 x\) b) \(g'(x)=-\csc^2 x\)

- Compare Tane's secant formula with the derivative family for reciprocal trigonometric functions. - One corrected formula contains tangent, not a second secant factor. - Check the sign of the cosecant derivative.

1. The secant derivative contains both secant and tangent: \((\sec x)'=\sec x\tan x\). 2. The cosecant derivative has a negative sign: \((\csc x)'=-\csc x\cot x\).

\((\sec x)'=\sec x\tan x\) \((\csc x)'=-\csc x\cot x\)

- You can rewrite cosecant as a quotient with numerator \(1\). - Try checking the result with the quotient rule, then write it in terms of reciprocal trigonometric functions.

1. The cosecant derivative is \(f'(x)=-\csc x\cot x\).

\(f'(x)=-\csc x\cot x\)

- Differentiate the trigonometric function first, taking care with its sign. - Then account for the constant multiplying that function.

1. The derivative of cotangent is \(-\csc^2x\). 2. The negative factor in \(g\) changes the sign, so \(g'(x)=\csc^2x\).

\(g'(x)=\csc^2x\)

- Differentiate the two trigonometric terms separately. - Keep each constant coefficient attached to its term. - Check the derivative formulas for both tangent and secant before combining them.

1. Differentiate each term using the tangent and secant derivative formulas. 2. The derivative is \(h'(x)=3\sec^2 x-2\sec x\tan x\).

\(h'(x)=3\sec^2 x-2\sec x\tan x\)

- Use the derivative formula for cotangent. - Evaluate the reciprocal of sine at the standard angle. - Preserve the negative sign from the derivative formula.

1. The derivative is \(f'(x)=-\csc^2 x\). 2. Since \(\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\), \(\csc\left(\frac{\pi}{4}\right)=\sqrt{2}\). 3. Therefore, \(f'\left(\frac{\pi}{4}\right)=-(\sqrt{2})^2=-2\).

\(-2\)

- Differentiate tangent and secant separately before substituting the angle. - Use the exact values of secant and tangent at \(\frac{\pi}{3}\). - Compare the two exact derivative values rather than converting immediately to decimals.

1. Differentiate: \(p'(x)=\sec^2x\) and \(q'(x)=\sec x\tan x\). 2. At \(x=\frac{\pi}{3}\), \(\sec\left(\frac{\pi}{3}\right)=2\) and \(\tan\left(\frac{\pi}{3}\right)=\sqrt{3}\). 3. Therefore, \(p'\left(\frac{\pi}{3}\right)=4\) and \(q'\left(\frac{\pi}{3}\right)=2\sqrt{3}\). 4. Since \(4>2\sqrt{3}\), \(p\) has the larger instantaneous rate by \(4-2\sqrt{3}\).

\(p\) changes faster. The rates are \(4\) and \(2\sqrt{3}\), so the difference is \(4-2\sqrt{3}\).

- Start with the derivative formula for secant. - Use a Pythagorean identity to determine tangent from the given secant value. - The quadrant tells you which sign to use for tangent.

1. The derivative is \(\frac{d}{dx}(\sec x)=\sec x\tan x\). 2. Use \(\tan^2\theta=\sec^2\theta-1\): \(\tan^2\theta=4-1=3\). 3. Because \(\theta\) is in Quadrant I, \(\tan\theta=\sqrt{3}\). 4. Therefore, \(\sec\theta\tan\theta=2\sqrt{3}\).

\(2\sqrt{3}\)

- Rewrite the derivative in terms of cosine to inspect its domain. - A reciprocal expression is undefined when its denominator is zero. - List only the relevant zeros of cosine in the stated interval.

1. The derivative is \(f'(x)=\sec^2 x=\frac{1}{\cos^2 x}\). 2. This expression is undefined where \(\cos x=0\). 3. On \([0,2\pi)\), those inputs are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

\(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). At these inputs, \(\cos x=0\), so both \(\tan x\) and its derivative are undefined.

- How might rewriting secant as a reciprocal help check its derivative rule? - Before evaluating a derivative at an input, check whether the original function is defined there.

1. The secant rule gives \(f'(x)=\sec x\tan x\). Hence \(f'(0)=1\cdot0=0\). 2. Since \(\sec x=\frac{1}{\cos x}\), the function is undefined at \(x=\frac{\pi}{2}\); its derivative cannot exist there.

\(f'(x)=\sec x\tan x\), so \(f'(0)=0\). The derivative at \(\frac{\pi}{2}\) does not exist because \(\cos(\frac{\pi}{2})=0\), so \(f\) is undefined there.

- Use the reciprocal relationship between cosecant and sine to check the derivative rule. - Think about which standard-angle values are exact rather than decimal approximations. - Does the function rise or fall near this angle? Use that to check your slope's sign.

1. The derivative is \(f'(x)=-\csc x\cot x\). 2. At \(x=\frac{\pi}{4}\), \(\csc x=\sqrt{2}\) and \(\cot x=1\). 3. The tangent slope is \(f'(\frac{\pi}{4})=-\sqrt{2}\).

\(-\sqrt{2}\)

- Differentiate cotangent and secant using different rules. - Use standard-angle values and check both derivative signs. - Compare the signed slopes rather than their absolute values.

1. The derivatives are \(p'(x)=-\csc^2x\) and \(q'(x)=\sec x\tan x\). 2. At \(\pi/3\), \(p'=-\frac{4}{3}\) and \(q'=2\sqrt{3}\). 3. Since \(2\sqrt{3}>-\frac{4}{3}\), \(q\) has the larger tangent slope.

\(p'(\frac{\pi}{3})=-\frac{4}{3}\), \(q'(\frac{\pi}{3})=2\sqrt{3}\). The secant function has the larger slope.

- Recall the two derivative rules and their signs. - Evaluate both terms at the same standard angle. - Watch the minus sign when combining the two derivative terms.

1. Differentiate both terms: \(f'(x)=\sec^2x-\csc^2x\). 2. At \(x=\frac{\pi}{4}\), both squared reciprocal functions equal \(2\). 3. Therefore, \(f'(\frac{\pi}{4})=2-2=0\).

\(f'(x)=\sec^2x-\csc^2x\); \(f'(\frac{\pi}{4})=0\).

- Recall the derivative of cosecant. - Compare the signs of sine and cosine at the two angles.

1. The cosecant rule gives \(f'(x)=-\csc x\cot x\). 2. At \(\frac{\pi}{6}\), cosecant is \(2\) and cotangent is \(\sqrt{3}\), so \(f'(\frac{\pi}{6})=-2\sqrt{3}\). 3. At \(\frac{5\pi}{6}\), cosecant is \(2\) and cotangent is \(-\sqrt{3}\), so \(f'(\frac{5\pi}{6})=2\sqrt{3}\). 4. Cotangent changes sign at these angles while cosecant stays positive, so the two slopes have opposite signs.

a) \(f'(\frac{\pi}{6})=-2\sqrt{3}\); \(f'(\frac{5\pi}{6})=2\sqrt{3}\). b) Cotangent has opposite signs at these angles, while cosecant is positive at both.

- Translate the prescribed tangent slope into an equation involving the derivative. - Rewrite secant in terms of cosine before solving. - Check all standard-angle solutions in the stated interval.

1. Since \(f'(x)=\sec^2 x\), the slope condition is \(\sec^2 x=4\). 2. This is equivalent to \(\cos^2 x=\frac{1}{4}\), so \(\cos x=\pm\frac{1}{2}\). 3. On \([0,\pi)\), the solutions are \(x=\frac{\pi}{3}\) and \(x=\frac{2\pi}{3}\).

\(x=\frac{\pi}{3}\) and \(x=\frac{2\pi}{3}\)

- Compare the sign of a derivative with whether the original graph is increasing or decreasing. - Use the slope of \(\tan x\) at the origin as a second identifying feature. - A function graph and its derivative need not share the same intercepts.

1. The derivative of \(\tan x\) must be positive throughout this interval because \(\tan x\) is increasing there. 2. At \(x=0\), the tangent slope of \(\tan x\) is \(1\), so the derivative graph must pass through \((0,1)\). 3. Graph q stays positive and passes through \((0,1)\), while graph p passes through the origin. Therefore, q is the derivative graph.

q is the derivative graph.

- Estimate the slope near the origin using a small rise and run. - How does the tangent derivative rule let you check the graphical estimate? - What happens to the original function near the dashed lines?

1. The curve rises about one unit for each unit of horizontal change close to the origin, so its tangent slope there is about \(1\). 2. By the tangent rule, \(f'(x)=\sec^2x\), giving \(f'(0)=1\). 3. Each dashed line is a vertical asymptote where \(\tan x\) is undefined. No derivative exists there.

From the graph, \(f'(0)\approx1\). By the rule, \(f'(x)=\sec^2x\), so \(f'(0)=1\). No derivative exists at either dashed vertical asymptote because \(f\) is undefined there.

- Use symmetry to distinguish the even secant graph from its derivative graph. - Derive the secant formula from \(1/\cos x\) rather than quoting it in part b). - For the parity argument, combine the parity of secant and tangent. - In part d), a substitution involving \(\tan x\) turns the equation into an algebraic equation in \(\tan^2 x\).

1. Graph p is positive and even, matching \(\sec x\). Graph q is therefore \(f'\). 2. Write \(f(x)=\frac{1}{\cos x}\). By the quotient rule, \(f'(x)=\frac{0\cdot\cos x-1(-\sin x)}{\cos^2 x}=\frac{\sin x}{\cos^2 x}=\sec x\tan x\). 3. Since \(\sec(-x)=\sec x\) and \(\tan(-x)=-\tan x\), we have \(f'(-x)=\sec(-x)\tan(-x)=-\sec x\tan x=-f'(x)\). Thus \(f'\) is odd. 4. On \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \(\sec x>0\). Let \(t=\tan x\). Then \(\sec x=\sqrt{1+t^2}\), so \(f'(x)=2\) becomes \(t\sqrt{1+t^2}=2\). Because the right side is positive, \(t>0\). 5. Squaring gives \(t^2(1+t^2)=4\). Let \(u=t^2\). Then \(u^2+u-4=0\), so \(u=\frac{-1+\sqrt{17}}{2}\). Hence \(t=\sqrt{\frac{-1+\sqrt{17}}{2}}\). 6. Therefore, \(x=\arctan\!\left(\sqrt{\frac{-1+\sqrt{17}}{2}}\right)\approx0.896\).

a) q is \(f'\). b) \(f'(x)=\sec x\tan x\) c) \(f'\) is odd because \(\sec x\) is even and \(\tan x\) is odd. d) \(x=\arctan\!\left(\sqrt{\frac{-1+\sqrt{17}}{2}}\right)\approx0.896\)