Free math worksheets

- Recall the derivative pair for tangent and cotangent. - Only one of these two formulas carries a negative sign.

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 the student'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: \((\csc x)'=-\csc x\cot x\).

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

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

- What derivative value corresponds to a horizontal tangent? - Use both the product rule and the chain rule. - Include the derivative of the inner expression \(c-x\). - Substitute \(x=\frac{c}{2}\) before doing unnecessary simplification.

1. Differentiate using the product and chain rules: \(f'(x)=\sec^2(x)\tan(c-x)-\tan(x)\sec^2(c-x)\). 2. Substitute \(x=\frac{c}{2}\): \(f'\left(\frac{c}{2}\right)=\sec^2\left(\frac{c}{2}\right)\tan\left(\frac{c}{2}\right)-\tan\left(\frac{c}{2}\right)\sec^2\left(\frac{c}{2}\right)=0\). 3. Therefore, the tangent slope is \(0\), so the graph has a horizontal tangent at \(x=\frac{c}{2}\). The condition that \(f\) is defined there is equivalent to \(\cos\left(\frac{c}{2}\right)\neq0\).

\(f'\left(\frac{c}{2}\right)=0\), so the tangent is horizontal whenever \(\cos\left(\frac{c}{2}\right)\neq0\).

- Determine the values of cosine on the stated interval. - Use point-slope form for the tangent line. - Find both axis intercepts of the tangent line. - Use the area formula for a right triangle.

1. On the given interval, \(0\leq\cos(x)\leq1\). Since \(k>0\), the range is \([0, k]\). 2. Since \(h_k\left(\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{2}\) and \(h_k'(x)=-k\sin(x)\), the tangent slope is \(-\frac{k\sqrt{2}}{2}\). Thus \(t_k: y=-\frac{k\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)+\frac{k\sqrt{2}}{2}\). 3. The y-intercept is \(\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\), and the x-intercept is \(1+\frac{\pi}{4}\). Therefore, \(A(k)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\). 4. Set \(A(k)=10\): \(\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2=10\). Solving gives \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\).

1. \([0, k]\) 2. \(t_k: y=-\frac{k\sqrt{2}}{2}x+\frac{k\sqrt{2}}{2}\left(1+\frac{\pi}{4}\right)\) 3. \(A(k)=\frac{k\sqrt{2}}{4}\left(1+\frac{\pi}{4}\right)^2\) 4. \(k=\frac{20\sqrt{2}}{\left(1+\frac{\pi}{4}\right)^2}\approx8.87\)

- 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.

- 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) \(x=\arctan\!\left(\sqrt{\frac{-1+\sqrt{17}}{2}}\right)\approx0.896\)