Differentiating Inverse Trigonometric Functions - AP Calculus AB

1. A tracking system defines $m(x)=\arccos(2-6x)$. What is $m'(x)$?
   Answer: $\dfrac{6}{\sqrt{1-(2-6x)^2}}$ This requires differentiating arccos with a decreasing linear argument. The derivative of arccos(u) is -1/√(1-u²), and for u = 2-6x, we apply the chain rule. Calculating: m'(x) = -1/√(1-(2-6x)²) · d/dx(2-6x) = -1/√(1-(2-6x)²) · (-6) = 6/√(1-(2-6x)²). Option B incorrectly shows a negative result, not recognizing that arccos's negative derivative times the negative derivative of 2-6x yields positive. Track both signs: arccos contributes negative, decreasing inner function contributes negative, producing positive overall.
2. A measurement model is $r(t)=\arcsin\!\left(\frac{t-4}{3}\right)$. What is $r'(t)$?
   Answer: $\dfrac{1}{3\sqrt{1-\left(\frac{t-4}{3}\right)^2}}$ This problem involves differentiating arcsin with a fractional linear argument. The derivative of arcsin(u) is 1/√(1-u²), and for u = (t-4)/3, we need the chain rule. Computing: r'(t) = 1/√(1-((t-4)/3)²) · d/dt((t-4)/3) = 1/√(1-((t-4)/3)²) · (1/3) = 1/(3√(1-((t-4)/3)²)). Option A omits the factor of 1/3 from differentiating the inner function. When differentiating expressions like (t-a)/b, the derivative is 1/b, which must be included.
3. In a motion model, $s(t)=\arctan(3t)$. What is $s'(t)$?
   Answer: $\dfrac{3}{1+9t^2}$ This problem requires differentiating an inverse trigonometric function, specifically arctangent. The derivative of arctan(u) is 1/(1+u²) times the derivative of u, applying the chain rule. Here, u = 3t, so we need to multiply by the derivative of 3t, which is 3. This gives us 3/(1+9t²), since (3t)² = 9t². A common error would be forgetting the chain rule and writing 1/(1+9t²), which incorrectly omits the factor of 3. Remember: when differentiating inverse trig functions with a composite argument, always multiply by the derivative of the inner function.
4. For $y=\arcsec(5x)$, what is $\dfrac{dy}{dx}$?
   Answer: $\dfrac{5}{|5x|\sqrt{(5x)^2-1}}$ This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse secant. The derivative of arcsec(u) is 1/(|u| √(u² - 1)) times du/dx. u = 5x, du/dx = 5, positive. Secant property. Result: 5 / (|5x| √((5x)² - 1)). Choice C negative, confuses with arccsc. Recognize arcsec positive with absolute.
5. For $y=\arcsin(\tfrac{3x+2}{5})$, what is $\dfrac{dy}{dx}$?
   Answer: $\dfrac{3}{5\sqrt{1-\left(\tfrac{3x+2}{5}\right)^2}}$ This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse sine. The derivative of arcsin(u) is 1/√(1 - u²) times du/dx. u = (3x + 2)/5, du/dx = 3/5, positive. Uses sine identity. Result: (3/5) / √(1 - ((3x + 2)/5)²). Choice A omits 5, wrong scaling. Pattern: arcsin positive with positive u'.
6. A calibration curve uses $u(x)=\arccot(2x)$. What is $u'(x)$?
   Answer: $-\dfrac{2}{1+(2x)^2}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccot function. The derivative of arccot(u) is -1 over 1 plus u squared, times u'. With u as 2x, u' is 2, so -2 over 1 plus (2x) squared. This negative form comes from arccot's inherent decrease. Choice A omits the negative, which is a common oversight. For patterns, always include the negative in arccot derivatives before chain rule multiplication.
7. A control system uses $h(x)=\arctan(5x+4)$. What is $h'(x)$?
   Answer: $\dfrac{5}{1+(5x+4)^2}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. The formula for the derivative of arctan(u) is 1 over 1 plus u squared, multiplied by u'. Here, u is 5x+4, so u' is 5, yielding 5 over 1 plus (5x+4) squared after chain rule application. This shows how the derivative measures sensitivity to changes in the linear argument. Choice A errs by placing the 5 in the numerator without squaring the denominator term correctly. Recognize arctan patterns by ensuring the chain rule multiplier matches the coefficient of x in the argument.
8. In a lab, $s(t)=\arctan(3t-1)$ models a sensor angle. What is $s'(t)$?
   Answer: $\dfrac{3}{1+(3t-1)^2}$ This problem requires differentiating an inverse trigonometric function, specifically arctan with a composite argument. The derivative of arctan(u) is 1/(1+u²), and we must apply the chain rule since u = 3t-1. Taking the derivative: s'(t) = 1/(1+(3t-1)²) · d/dt(3t-1) = 1/(1+(3t-1)²) · 3 = 3/(1+(3t-1)²). A common error would be forgetting the chain rule and selecting option B, which omits the factor of 3 from the inner derivative. Remember: when differentiating inverse trig functions with composite arguments, always multiply by the derivative of the inner function.
9. An angle function is $b(x)=\arcsin(2-\tfrac{x}{3})$. What is $b'(x)$?
   Answer: $-\dfrac{1}{3}{\sqrt{1-(2-\tfrac{x}{3})^2}}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsin function. Derivative of $\arcsin(u)$ with u as $2 - \tfrac{x}{3}$ is $\frac{1}{\sqrt{1 - u^2}}$ times $-\frac{1}{3}$. This results in $-\frac{1}{3}$ over the sqrt term, sign from decreasing u. It adjusts for the negative slope in the argument. Choice A uses positive $\frac{1}{3}$, ignoring u's derivative sign. Pattern: Negative coefficients in arcsin arguments lead to negative derivatives.
10. Let $g(x)=\arcsin(\tfrac{5-x}{2})$. What is $g'(x)$?
    Answer: $-\dfrac{1/2}{\sqrt{1-\left(\tfrac{5-x}{2}\right)^2}}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsin function. For arcsin(u) with u as (5-x)/2, derivative is 1 over sqrt(1 minus u squared) times -1/2. Yields -1/2 over the sqrt term, negative from decreasing u. Adjusts for the scaled, decreasing argument. Choice A uses positive 1/2, ignoring sign. Pattern: Decreasing scaled arcsin gives negative fractional derivative.
11. Let $j(x)=\arccsc(2x-5)$. What is $j'(x)$?
    Answer: $-\dfrac{2}{|2x-5|\sqrt{(2x-5)^2-1}}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccsc function. For arccsc(u) with u as 2x-5, derivative is -1 over |u| sqrt(u squared minus 1) times 2. Results in -2 over |2x-5| sqrt(...). Combines negative with scaling. Choice B omits negative, mistaking for arcsec. Pattern: Arccsc with positive coefficient keeps overall negative after multiplication.
12. Given $d(x)=\arccot(4x-3)$, what is $d'(x)$?
    Answer: $-\dfrac{4}{1+(4x-3)^2}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccot function. Derivative of arccot(u) with u as 4x-3 is $-\frac{1}{1 + u^2} \times 4$. Results in $-\frac{4}{1 + (4x-3)^2}$. The negative is inherent, scaled up by 4. Choice A forgets the negative, resembling arctan. Pattern: Arccot always starts negative, then multiply by u'.
13. For a parameter study, $p(x)=\arctan(x^2+1)$. What is $p'(x)$?
    Answer: $\dfrac{2x}{1+(x^2+1)^2}$ This question requires differentiating arctangent of a quadratic expression. The derivative of arctan(u) is 1/(1+u²) times the derivative of u. Here, u = x²+1, so we multiply by 2x (the derivative of x²+1). This gives us 2x/(1+(x²+1)²), where the entire expression (x²+1) is squared in the denominator. A common error would be to expand (x²+1)² prematurely or to use a square root in the denominator like 2x/√(1+(x²+1)²). For arctan derivatives, remember: no square root in the denominator, and keep composite expressions intact when squaring.
14. A sensor output is $q(t)=\arcsin(1+t^2)$. What is $q'(t)$?
    Answer: $\dfrac{2t}{\sqrt{1-(1+t^2)^2}}$ This problem involves differentiating arcsin of a quadratic expression. The derivative of arcsin(u) is 1/√(1-u²) times the derivative of u. With u = 1+t², we multiply by 2t (the derivative of 1+t²). This yields 2t/√(1-(1+t²)²), keeping (1+t²) as a unit when squaring. A student might incorrectly write 2t/(1-(1+t²)²) without the square root, confusing arcsin with arctan formulas. When differentiating inverse sine or cosine, always include the square root in the denominator—this is what distinguishes them from inverse tangent.
15. In a geometry app, $r(x)=\arcsin(x-6)$. What is $r'(x)$?
    Answer: $\dfrac{1}{\sqrt{1-(x-6)^2}}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsin function. For arcsin(u) where u is x-6, the derivative is 1 over sqrt(1 minus u squared) times u', which is 1. This gives 1 over sqrt(1 minus (x-6) squared), applying the chain rule implicitly with u' as 1. It reflects the rate at which the inverse sine changes with a shifted input. Choice A incorrectly omits the squared term in the square root, altering the domain. For pattern recognition, treat shifted arguments in arcsin by directly substituting into the formula without extra adjustments beyond chain rule.
16. A formula uses $f(x)=\arctan(\tfrac{3}{2}x)$. What is $f'(x)$?
    Answer: $\dfrac{3/2}{1+\left(\tfrac{3}{2}x\right)^2}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. Derivative of arctan(u) with u as (3/2)x is 1 over 1 plus u squared times 3/2. Results in 3/2 over 1 plus ((3/2)x) squared. Increases magnitude from the 3/2 factor. Choice B uses 3 instead, doubling incorrectly. Pattern: Multiply standard arctan derivative by the fractional coefficient.
17. In a modeling context, $g(t)=\arcsin(7-t)$. What is $g'(t)$?
    Answer: $\dfrac{-1}{\sqrt{1-(7-t)^2}}$ This question requires differentiating arcsin with a decreasing linear argument. The derivative of arcsin(u) is 1/√(1-u²) times the derivative of u. Since u = 7-t, the derivative of u is -1. This gives us -1/√(1-(7-t)²), where we keep (7-t) intact and squared under the radical. A common mistake would be to write 1/√(1-(7-t)²), forgetting that the derivative of 7-t is negative. Always pay attention to the sign when differentiating expressions like a-x, which have negative derivatives.
18. A function is $s(x)=\arcsec(\tfrac{3x+1}{2})$. What is $s'(x)$?
    Answer: $\dfrac{3/2}{\left|\tfrac{3x+1}{2}\right|\sqrt{\left(\tfrac{3x+1}{2}\right)^2-1}}$ This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse secant. The derivative of arcsec(u) is 1/(|u| √(u² - 1)) times du/dx. Here, u = (3x + 1)/2, du/dx = 3/2, yielding positive. This formula reflects secant's reciprocal nature. Thus, it's (3/2) / (|(3x + 1)/2| √(((3x + 1)/2)² - 1)). Choice C adds an extra negative, perhaps confusing with arccsc. To recognize, look for arcsec with absolute values and positive base adjusted by chain rule.
19. A signal is modeled by $y(x)=\arcsin(2x)$. What is $y'(x)$?
    Answer: $\dfrac{2}{\sqrt{1-4x^2}}$ This problem involves differentiating the inverse sine function with a composite argument. The derivative of arcsin(u) is 1/√(1-u²) times the derivative of u. Since u = 2x, we multiply by 2 (the derivative of 2x). This yields 2/√(1-4x²), where we substitute (2x)² = 4x² in the denominator. A tempting mistake would be to write 1/√(1-4x²), forgetting to apply the chain rule and multiply by 2. When differentiating inverse trig functions, always check if the argument is composite and apply the chain rule accordingly.
20. A function is $h(x)=\arccos(1-9x)$. What is $h'(x)$?
    Answer: $\dfrac{9}{\sqrt{1-(1-9x)^2}}$ This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccos function. Derivative of arccos(u) with u as 1-9x is -1 over sqrt(1 minus u squared) times -9. Negatives cancel to 9 over the sqrt term. Shows sign flip from chain rule. Choice A adds unnecessary negative, forgetting cancellation. Pattern: Large negative coefficients in arccos can yield positive large multipliers.

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