A motion model is on its domain; what is ?
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AP Calculus BC Quiz
Practice Derivatives Of Trigonometry And Logarithmic Functions in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A motion model is h(x)=ln(cosx) on its domain; what is h′(x)?
This quiz focuses on Derivatives Of Trigonometry And Logarithmic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.
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.
A motion model is h(x)=ln(cosx) on its domain; what is h′(x)?
Explanation: This problem tests the skill of differentiating logarithmic functions composed with trigonometric functions. The function is h(x) = ln(cos x), requiring the chain rule for the composition. The derivative of ln(u) is (1/u) u', where u = cos x and u' = -sin x. Thus, h'(x) = (1/cos x) * (-sin x) = -sin x / cos x. This simplifies to -tan x, but the form matches the choice. A tempting distractor is A, 1/cos x, which forgets the chain rule's multiplication by u' = -sin x. A transferable strategy is to apply the chain rule for ln(u(x)) by computing (1/u) times u', always including the inner derivative.
A signal is modeled by f(x)=3sinx−2ex. What is f′(x)?
Explanation: This problem involves differentiating trigonometric and exponential functions. The derivative of sin(x) is cos(x), so the derivative of 3sin(x) is 3cos(x). The derivative of e^x is e^x, so the derivative of -2e^x is -2e^x. Therefore, f'(x) = 3cos(x) - 2e^x. Students might confuse the derivative of sin(x) with -sin(x), which would lead to the incorrect answer C. When differentiating, remember that d/dx[sin(x)] = cos(x) and d/dx[e^x] = e^x.
The function g(x)=5lnx+4sinx models a rate. What is g′(x)?
Explanation: This problem requires differentiating logarithmic and trigonometric functions. The derivative of ln(x) is 1/x, so the derivative of 5ln(x) is 5/x. The derivative of sin(x) is cos(x), so the derivative of 4sin(x) is 4cos(x). Therefore, g'(x) = 5/x + 4cos(x). A common mistake would be to confuse the derivative of sin(x) with sin(x) itself, leading to answer C. Remember the fundamental derivatives: d/dx[ln(x)] = 1/x and d/dx[sin(x)] = cos(x).
The cost is C(x)=lnx−6cosx. What is C′(x) for x>0?
Explanation: This problem requires differentiating logarithmic and trigonometric functions. The derivative of ln(x) is 1/x. For the term -6cos(x), the derivative of cos(x) is -sin(x), so the derivative of -6cos(x) is -6(-sin(x)) = 6sin(x). Therefore, C'(x) = 1/x + 6sin(x). A common error would be to forget the negative sign when differentiating cos(x), leading to answer A with -6cos(x). Remember that d/dx[cos(x)] = -sin(x), and when combined with a negative coefficient, the result becomes positive.
A particle’s position is s(t)=4sint−3t2. What is s′(t)?
Explanation: This problem requires differentiating a combination of trigonometric and polynomial functions. The derivative of sin(t) is cos(t), so the derivative of 4sin(t) is 4cos(t). For the polynomial term -3t², we apply the power rule: the derivative is -3(2t) = -6t. Combining these results, s'(t) = 4cos(t) - 6t. A common error would be to think the derivative of sin(t) is -sin(t), which would incorrectly lead to answer C. When differentiating, remember that d/dx[sin(x)] = cos(x) and apply the constant multiple rule to handle coefficients.
A chemical concentration is C(t)=ln(4t)+7. What is C′(t)?
Explanation: This question tests the skill of differentiating basic logarithmic functions. The derivative of ln(kt), where k is a constant, is (1/(kt)) * k = 1/t, simplifying due to the chain rule. For C(t) = ln(4t) + 7, the derivative is 1/t, and the constant +7 has a derivative of zero. Therefore, C'(t) = 1/t. A tempting distractor is choice A, 1/(4t), which forgets to apply the chain rule by multiplying by the derivative of 4t, which is 4. When differentiating natural logarithms of linear functions, apply the chain rule to account for the inner derivative.
A moving object position is p(t)=sint+et+lnt. What is p′(t)?
Explanation: This question tests the skill of differentiating combinations of trigonometric, exponential, and logarithmic functions. The derivative of sin t is cos t, of e^t is e^t, and of ln t is 1/t. For p(t) = sin t + e^t + ln t, the result is cos t + e^t + 1/t. Therefore, p'(t) = cos t + e^t + 1/t. A tempting distractor is choice C, -cos t + e^t + 1/t, which incorrectly uses the derivative of cosine instead of sine. For sums of diverse functions, differentiate each component using its fundamental rule and sum the results.
A signal is modeled by s(t)=5sin(3t)+2. What is s′(t)?
Explanation: This question tests the skill of differentiating basic trigonometric functions. The derivative of a sine function of the form a sin(kt) is a k cos(kt), where a is a constant and k is the coefficient inside the sine. For s(t) = 5 sin(3t) + 2, the derivative is 5 * 3 cos(3t) = 15 cos(3t), and the constant +2 differentiates to zero. Therefore, s'(t) = 15 cos(3t). A tempting distractor is choice A, 5 cos(3t) + 2, which neglects the chain rule multiplication by 3 and incorrectly keeps the constant. To differentiate trigonometric functions with linear arguments, remember to use the chain rule and multiply by the inner derivative.
A population model is P(t)=5lnt−2et. What is P′(t) for t>0?
Explanation: This question tests basic function differentiation with logarithmic and exponential functions. To find P'(t), we differentiate each term: the derivative of 5ln(t) is 5·(1/t) = 5/t, and the derivative of -2e^t is -2e^t. Therefore, P'(t) = 5/t - 2e^t. A tempting error would be to confuse the derivative of ln(t) with the function itself, perhaps writing 5ln(t) instead of 5/t. When differentiating functions involving natural logarithms, remember that d/dx[ln(x)] = 1/x.
A periodic displacement is d(x)=sin(πx). What is d′(x)?
Explanation: This question tests basic function differentiation with trigonometric functions using chain rule. To find d'(x), we differentiate sin(πx): the derivative is cos(πx) multiplied by the derivative of πx, which is π. Therefore, d'(x) = π·cos(πx). A tempting error would be to forget the chain rule factor π, yielding just cos(πx). When differentiating sin(ax) where a is a constant, multiply by that constant.
A particle’s position is s(t)=4sint−3cost. What is s′(t)?
Explanation: This question tests basic function differentiation using trigonometric derivatives. To find s'(t), we differentiate each term: the derivative of 4sin(t) is 4cos(t), and the derivative of -3cos(t) is -3(-sin(t)) = 3sin(t). Therefore, s'(t) = 4cos(t) + 3sin(t). A common error would be forgetting that the derivative of cosine is negative sine, leading to the wrong sign on the second term. When differentiating sums of basic functions, apply the derivative rules to each term independently and combine the results.
In acoustics, B(x)=sinx−lnx. What is B′(x) for x>0?
Explanation: This question involves basic function differentiation combining trigonometric and logarithmic functions. To find B'(x), we differentiate each term: the derivative of sin(x) is cos(x), and the derivative of -ln(x) is -1/x. Therefore, B'(x) = cos(x) - 1/x. A common mistake would be to forget the negative sign on the second term or confuse the derivatives of sine and cosine. When differentiating differences of functions, preserve the signs and apply derivative rules to each term.
In chemistry, k(t)=et+ln(t2). What is k′(t) for t=0?
Explanation: This question involves basic function differentiation with exponential and logarithmic functions using properties of logarithms. To find k'(t), we differentiate each term: the derivative of e^t is e^t, and for ln(t²) = 2ln(t), the derivative is 2/t. Therefore, k'(t) = e^t + 2/t. A common error would be to write 1/t instead of 2/t for the derivative of ln(t²). When differentiating ln(x^n), remember that the coefficient n appears in the numerator of the derivative.
A waveform is w(x)=3cosx−2sinx. What is w′(x)?
Explanation: This question tests basic function differentiation with trigonometric functions. To find w'(x), we differentiate each term: the derivative of 3cos(x) is 3(-sin(x)) = -3sin(x), and the derivative of -2sin(x) is -2cos(x). Therefore, w'(x) = -3sin(x) - 2cos(x). A tempting mistake would be to get the signs wrong, particularly forgetting that the derivative of cosine is negative sine. When differentiating linear combinations of sine and cosine, carefully apply the sign rules.
In optics, n(x)=ln(x2). What is n′(x) for x=0?
Explanation: This question involves basic function differentiation with logarithmic functions using properties of logarithms. To find n'(x), we first note that ln(x²) = 2ln(x), so n(x) = 2ln(x). The derivative is then 2·(1/x) = 2/x. Alternatively, using chain rule directly: the derivative of ln(x²) is 1/x² multiplied by 2x, giving 2x/x² = 2/x. A common error would be to write 1/x² without applying the chain rule. When differentiating ln(x^n), the result is n/x.
A process is modeled by f(t)=cost−e−t. What is f′(t)?
Explanation: This question involves basic function differentiation combining trigonometric and exponential functions. To find f′(t), we differentiate each term: the derivative of cos(t) is −sin(t), and the derivative of −e−t is −(e−t⋅(−1))=e−t. Therefore, f′(t)=−sin(t)+e−t. A common error would be to get the sign wrong on either term, particularly when differentiating −e−t. When differentiating −e−x, apply the chain rule: the result is +e−x.
A measurement model is M(x)=ln(5x)−cosx. What is M′(x) for x>0?
Explanation: This question involves basic function differentiation with logarithmic and trigonometric functions using chain rule. To find M′(x), we differentiate each term: the derivative of ln(5x) is 5x1⋅5=x1, and the derivative of −cos(x) is −(−sin(x))=sin(x). Therefore, M′(x)=x1+sin(x). A common mistake would be to write 5x1 without applying the chain rule properly. When differentiating ln(ax), the result simplifies to x1 regardless of the constant a.
A log model is L(t)=ln(et). What is L′(t)?
Explanation: This question involves basic function differentiation with logarithmic functions using properties of logarithms. To find L'(t), we note that ln(e^t) = t by the property that ln and e are inverse functions. Therefore, L(t) = t, and L'(t) = 1. A common error would be to apply the chain rule unnecessarily, writing e^t or 1/t. When differentiating ln(e^x), remember that these functions are inverses, so the result simplifies to 1.
A decay model is m(t)=9e−t+cost. What is m′(t)?
Explanation: This question tests the skill of differentiating combinations of exponential and trigonometric functions. The derivative of a e^{-t} is a (-1) e^{-t} = -a e^{-t}, and the derivative of cos t is -sin t. For m(t) = 9 e^{-t} + cos t, combine to get -9 e^{-t} - sin t. Thus, m'(t) = -9 e^{-t} - sin t. A tempting distractor is choice B, 9 e^{-t} - sin t, which forgets the negative sign from the exponential's chain rule. Always apply the chain rule correctly for exponentials and include signs for trigonometric derivatives.
A signal is f(x)=sin(3x)+lnx. What is f′(x) for x>0?
Explanation: This question involves basic function differentiation using trigonometric derivatives with chain rule and logarithmic functions. To find f'(x), we differentiate each term: the derivative of sin(3x) is cos(3x)·3 = 3cos(3x) by the chain rule, and the derivative of ln(x) is 1/x. Therefore, f'(x) = 3cos(3x) + 1/x. A common mistake would be to forget the factor of 3 when differentiating sin(3x), yielding just cos(3x). When differentiating composite trigonometric functions, always multiply by the derivative of the inner function.