During a test, gives an angle reading. What is ?
Opening subject page...
Loading your content
AP Calculus AB Quiz
Practice Differentiating Inverse Trigonometric Functions in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
Question 1 / 20
0 of 20 answered
During a test, f(x)=arccos(4x) gives an angle reading. What is f′(x)?
This quiz focuses on Differentiating Inverse Trigonometric Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
During a test, f(x)=arccos(4x) gives an angle reading. What is f′(x)?
Explanation: This requires differentiating arccos with a composite argument. The derivative of arccos(u) is -1/√(1-u²), noting the negative sign that distinguishes it from arcsin. With u = 4x and the chain rule: f'(x) = -1/√(1-(4x)²) · d/dx(4x) = -1/√(1-(4x)²) · 4 = -4/√(1-(4x)²). Option A incorrectly has a positive sign, forgetting that arccos has a negative derivative unlike arcsin. Remember the mnemonic: arccos goes down (negative derivative) while arcsin goes up (positive derivative).
A measurement model is r(t)=arcsin(3t−4). What is r′(t)?
Explanation: 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.
A signal phase shift is h(x)=arcsin(7−3x). What is h′(x)?
Explanation: This requires differentiating arcsin with a decreasing linear argument. The derivative of arcsin(u) is 1/√(1-u²), and for u = 7-3x, we need the chain rule. Calculating: h'(x) = 1/√(1-(7-3x)²) · d/dx(7-3x) = 1/√(1-(7-3x)²) · (-3) = -3/√(1-(7-3x)²). Option A incorrectly shows a positive result, forgetting that d/dx(7-3x) = -3 is negative. Always track the sign of the inner function's derivative—decreasing functions contribute a negative sign.
A controller uses q(x)=arctan(2x). What is q′(x)?
Explanation: This requires differentiating arctan with a fractional argument. The derivative of arctan(u) is 1/(1+u²), and for u = x/2, we apply the chain rule. Calculating: q'(x) = 1/(1+(x/2)²) · d/dx(x/2) = 1/(1+(x/2)²) · (1/2) = 1/(2(1+(x/2)²)). Option A forgets to include the factor of 1/2 from the chain rule, a common error with fractional arguments. Remember: when the inner function is x/a, its derivative is 1/a, which must multiply the outer derivative.
A sensor output is q(t)=arcsin(1+t2). What is q′(t)?
Explanation: 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.
A tracking function is v(x)=\arcsec(3x−2). What is v′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsec function. For arcsec(u) with u as 3x−2, the derivative is ∣u∣u2−11×u′, which is 3. Thus, ∣3x−2∣(3x−2)2−13. The absolute value maintains positivity in the derivative's structure. Choice A lacks the multiplier 3 from chain rule. Identify arcsec with linear u by scaling the standard form by u's coefficient.
For a signal phase shift, p(x)=\arcsec(x). What is p′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsec function. The derivative of arcsec(u) is 1 over |u| times the square root of u squared minus 1, multiplied by u'. Since u is x and u' is 1, it simplifies to 1 over |x| sqrt(x squared minus 1). The absolute value ensures the derivative's sign consistency with arcsec's domain. Choice A lacks the absolute value, which is crucial for correctness in all domains. Spot arcsec patterns by always including the absolute value in the denominator for generality.
Let p(x)=arccos(94−2x). What is p′(x)?
Explanation: This question tests the skill of differentiating inverse trigonometric functions, specifically the inverse cosine. The derivative of arccos(u) is -1/√(1 - u²) times du/dx. With u = (4 - 2x)/9 and du/dx = -2/9, the negative signs cancel to give a positive result. Conceptually, this reflects arccos being a decreasing function. The result is 2/(9 √(1 - ((4 - 2x)/9)²)). Choice B errs by including an unnecessary negative sign, perhaps from not accounting for the chain rule properly. To spot these, note arccos with decreasing inner functions often yields positive derivatives via chain rule.
Given d(x)=\arccot(4x−3), what is d′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccot function. Derivative of arccot(u) with u as 4x-3 is −1+u21×4. Results in −1+(4x−3)24. The negative is inherent, scaled up by 4. Choice A forgets the negative, resembling arctan. Pattern: Arccot always starts negative, then multiply by u'.
A model defines c(x)=arccos(5x−1). What is c′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccos function. For arccos(u) with u as (x-1)/5, derivative is -1 over sqrt(1 minus u squared) times 1/5. Yields -1/5 over the sqrt term, combining negative and scaling. Reflects reduced sensitivity from division by 5. Choice A omits the negative, mistaking for arcsin. Identify scaled arccos by including negative and dividing by the denominator.
For z(x)=arctan(21x−5), what is z′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. Derivative of arctan(u) with u as (1/2)x -5 is 1 over 1 plus u squared times 1/2. Result is 1/2 over 1 plus ((1/2)x -5) squared. This scales down the derivative due to the half coefficient. Choice D adds an unnecessary negative, confusing with arccot. Spot patterns in arctan with fractional coefficients by halving the standard derivative form.
A function is e(x)=\arcsec(2x+1). What is e′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsec function. For arcsec(u) with u as x/2 +1, derivative is 1 over |u| sqrt(u squared minus 1) times 1/2. Gives 1/2 over |x/2 +1| sqrt(...). Scales down due to half coefficient. Choice A lacks the 1/2, overlooking chain rule. Recognize fractional shifts in arcsec by halving the standard form.
A function is defined by a(x)=\arccsc(x+2). What is a′(x)?
Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccsc function. For \arccsc(u) with u as x+2, derivative is −∣u∣u2−11×1. Thus, −∣x+2∣(x+2)2−11. The negative is key to arccsc's behavior. Choice A lacks the negative sign, a frequent error. Recognize arccsc shifts by keeping the negative and absolute value intact.
A calibration curve uses u(x)=\arccot(2x). What is u′(x)?
Explanation: 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.
In a lab, s(t)=arctan(3t−1) models a sensor angle. What is s′(t)?
Explanation: 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.
In navigation, p(u)=arccos(1−u) represents a heading correction. What is p′(u)?
Explanation: This problem involves differentiating arccos with argument 1-u. The derivative of arccos(u) is -1/√(1-u²), and with u = 1-u (note the variable), we apply the chain rule. Computing: p'(u) = -1/√(1-(1-u)²) · d/du(1-u) = -1/√(1-(1-u)²) · (-1) = 1/√(1-(1-u)²). Option B incorrectly keeps the negative sign, not recognizing that two negatives make a positive. When arccos (negative derivative) contains a decreasing function (negative derivative), the result is positive.
A control system uses h(x)=arctan(5x+4). What is h′(x)?
Explanation: 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.
A signal is modeled by y(x)=arcsin(2x). What is y′(x)?
Explanation: 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.
In a modeling context, g(t)=arcsin(7−t). What is g′(t)?
Explanation: 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.
A measured angle satisfies h(x)=arccos(x2). What is h′(x)?
Explanation: This problem involves differentiating arccos of a power function. The derivative of arccos(u) is -1/√(1-u²) times the derivative of u. With u = x², we multiply by 2x (the derivative of x²). This yields -2x/√(1-x⁴), where (x²)² = x⁴ appears under the radical. A student might write -2x/(1-x⁴) without the square root, mixing up the formulas for arccos and arctan derivatives. Remember: arcsin and arccos derivatives always have square roots in their denominators, distinguishing them from arctan.