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AP Calculus AB

AP Calculus AB Quiz: Derivatives Of Trigonometry And Logarithmic Functions

Practice Derivatives Of Trigonometry And Logarithmic 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.

What this quiz covers

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 AB.

How to use this quiz

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.

Question 1

A function is D(x)=cos⁡ ⁣(sin⁡(2x))D(x)=\cos\!\left(\sin(2x)\right)D(x)=cos(sin(2x)). What is D′(x)D'(x)D′(x)?

  1. −sin⁡ ⁣(sin⁡(2x))-\sin\!\left(\sin(2x)\right)−sin(sin(2x))
  2. −2cos⁡(2x)sin⁡ ⁣(sin⁡(2x))-2\cos(2x)\sin\!\left(\sin(2x)\right)−2cos(2x)sin(sin(2x))
  3. 2cos⁡(2x)sin⁡ ⁣(sin⁡(2x))2\cos(2x)\sin\!\left(\sin(2x)\right)2cos(2x)sin(sin(2x))
  4. −sin⁡(2x)sin⁡ ⁣(sin⁡(2x))-\sin(2x)\sin\!\left(\sin(2x)\right)−sin(2x)sin(sin(2x))
  5. −2sin⁡(2x)sin⁡ ⁣(sin⁡(2x))-2\sin(2x)\sin\!\left(\sin(2x)\right)−2sin(2x)sin(sin(2x))
Explanation: Differentiating D(x) = cos(sin(2x)) requires nested chain rule. Outer cos(g), derivative -sin(g), g = sin(2x). g' = cos(2x) * 2. So D'(x) = -sin(g) * 2 cos(2x). A common omission is forgetting the 2 from the inner argument. Pattern: Spot double trig compositions and apply chain twice, outer to inner.

Question 2

For y(x)=cos⁡ ⁣((x2−1)4)y(x)=\cos\!\left((x^2-1)^4\right)y(x)=cos((x2−1)4), what is y′(x)y'(x)y′(x)?

  1. −sin⁡ ⁣((x2−1)4)-\sin\!\left((x^2-1)^4\right)−sin((x2−1)4)
  2. −8x(x2−1)3sin⁡ ⁣((x2−1)4)-8x(x^2-1)^3\sin\!\left((x^2-1)^4\right)−8x(x2−1)3sin((x2−1)4)
  3. −4(x2−1)3sin⁡ ⁣((x2−1)4)-4(x^2-1)^3\sin\!\left((x^2-1)^4\right)−4(x2−1)3sin((x2−1)4)
  4. 8x(x2−1)3cos⁡ ⁣((x2−1)4)8x(x^2-1)^3\cos\!\left((x^2-1)^4\right)8x(x2−1)3cos((x2−1)4)
  5. −8(x2−1)3sin⁡ ⁣((x2−1)4)-8(x^2-1)^3\sin\!\left((x^2-1)^4\right)−8(x2−1)3sin((x2−1)4)
Explanation: Cosine of a power function requires multiple chain rule applications. Outer cos⁡(u)\cos(u)cos(u), u=(x2−1)4u = (x^2 - 1)^4u=(x2−1)4, derivative −sin⁡(u)-\sin(u)−sin(u), inner u′=4(x2−1)3⋅2x=8x(x2−1)3u' = 4(x^2 - 1)^3 \cdot 2x = 8x(x^2 - 1)^3u′=4(x2−1)3⋅2x=8x(x2−1)3. y′(x)=−sin⁡((x2−1)4)⋅8x(x2−1)3y'(x) = -\sin((x^2 - 1)^4) \cdot 8x(x^2 - 1)^3y′(x)=−sin((x2−1)4)⋅8x(x2−1)3. Common omission: forgetting the 2x from inside the power. Missing the 4 from exponent or negative sign. Recognize trig of power of quadratic and layer the chain rule. This strategy applies to deep nests, ensuring thorough derivatives.

Question 3

A rate is given by u(x)=tan⁡ ⁣(ln⁡(x2+4))u(x)=\tan\!\left(\ln(x^2+4)\right)u(x)=tan(ln(x2+4)). What is u′(x)u'(x)u′(x)?

  1. sec⁡2 ⁣(ln⁡(x2+4))\sec^2\!\left(\ln(x^2+4)\right)sec2(ln(x2+4))
  2. 2xx2+4sec⁡2 ⁣(ln⁡(x2+4))\dfrac{2x}{x^2+4}\sec^2\!\left(\ln(x^2+4)\right)x2+42x​sec2(ln(x2+4))
  3. 1x2+4sec⁡2 ⁣(ln⁡(x2+4))\dfrac{1}{x^2+4}\sec^2\!\left(\ln(x^2+4)\right)x2+41​sec2(ln(x2+4))
  4. 2xx2+4tan⁡ ⁣(ln⁡(x2+4))\dfrac{2x}{x^2+4}\tan\!\left(\ln(x^2+4)\right)x2+42x​tan(ln(x2+4))
  5. 2xsec⁡2 ⁣(ln⁡(x2+4))2x\sec^2\!\left(\ln(x^2+4)\right)2xsec2(ln(x2+4))
Explanation: This tangent of a log requires chain rule for trig-log composite. Outer tan(u), u = ln(x^2 + 4), derivative sec^2(u), inner u' = 2x/(x^2 + 4). So u'(x) = (2x/(x^2 + 4)) sec^2(ln(x^2 + 4)). Common omission: forgetting 2x from inner derivative. Confusing tan derivative with sec tan is another error. Recognize trig of log polynomial and chain fully. This pattern aids in complex nests, building derivative skills.

Question 4

Let v(x)=ln⁡ ⁣(1−sin⁡x)v(x)=\ln\!\left(\sqrt{1-\sin x}\right)v(x)=ln(1−sinx​). What is v′(x)v'(x)v′(x)?

  1. 11−sin⁡x\dfrac{1}{\sqrt{1-\sin x}}1−sinx​1​
  2. −cos⁡x1−sin⁡x-\dfrac{\cos x}{\sqrt{1-\sin x}}−1−sinx​cosx​
  3. −cos⁡x2(1−sin⁡x)-\dfrac{\cos x}{2(1-\sin x)}−2(1−sinx)cosx​
  4. cos⁡x2(1−sin⁡x)\dfrac{\cos x}{2(1-\sin x)}2(1−sinx)cosx​
  5. ln⁡ ⁣(1−sin⁡x)⋅(−cos⁡x21−sin⁡x)\ln\!\left(\sqrt{1-\sin x}\right)\cdot \left(-\dfrac{\cos x}{2\sqrt{1-\sin x}}\right)ln(1−sinx​)⋅(−21−sinx​cosx​)
Explanation: Differentiating log of square root uses chain rule after simplification. v(x)=12ln⁡(1−sin⁡x)v(x) = \frac{1}{2} \ln(1 - \sin x)v(x)=21​ln(1−sinx), v′(x)=12⋅−cos⁡x1−sin⁡xv'(x) = \frac{1}{2} \cdot \frac{-\cos x}{1 - \sin x}v′(x)=21​⋅1−sinx−cosx​. Outer ln, inner 1−sin⁡x\sqrt{1 - \sin x}1−sinx​, but simplify first. Common omission: missing the 1/2 factor. Forgetting negative from -sin x derivative. Spot log of sqrt trig and use properties then chain. This transfers to radical logs, improving accuracy.

Question 5

Let G(x)=tan⁡ ⁣(1ln⁡x)G(x)=\tan\!\left(\dfrac{1}{\ln x}\right)G(x)=tan(lnx1​). What is G′(x)G'(x)G′(x)?

  1. sec⁡2 ⁣(1ln⁡x)\sec^2\!\left(\dfrac{1}{\ln x}\right)sec2(lnx1​)
  2. −1x(ln⁡x)2sec⁡2 ⁣(1ln⁡x)-\dfrac{1}{x(\ln x)^2}\sec^2\!\left(\dfrac{1}{\ln x}\right)−x(lnx)21​sec2(lnx1​)
  3. 1xln⁡xsec⁡2 ⁣(1ln⁡x)\dfrac{1}{x\ln x}\sec^2\!\left(\dfrac{1}{\ln x}\right)xlnx1​sec2(lnx1​)
  4. −1(ln⁡x)2sec⁡2 ⁣(1ln⁡x)-\dfrac{1}{(\ln x)^2}\sec^2\!\left(\dfrac{1}{\ln x}\right)−(lnx)21​sec2(lnx1​)
  5. −1x(ln⁡x)2tan⁡ ⁣(1ln⁡x)-\dfrac{1}{x(\ln x)^2}\tan\!\left(\dfrac{1}{\ln x}\right)−x(lnx)21​tan(lnx1​)
Explanation: For G(x) = tan(1 / ln x), chain rule applies. Outer tan(j), derivative sec²(j), j = (ln x)^{-1}. j' = - (ln x)^{-2} * (1/x) = -1/(x (ln x)²). Thus, G'(x) = sec²(j) * (-1/(x (ln x)²)). A common omission is sign error in the reciprocal derivative. Recognize trig of inverse log and use power rule in chain.

Question 6

For q(x)=ln⁡ ⁣(sin⁡(x3))q(x)=\ln\!\left(\sin(x^3)\right)q(x)=ln(sin(x3)), what is q′(x)q'(x)q′(x)?

  1. cos⁡(x3)sin⁡(x3)\dfrac{\cos(x^3)}{\sin(x^3)}sin(x3)cos(x3)​
  2. 1sin⁡(x3)\dfrac{1}{\sin(x^3)}sin(x3)1​
  3. 3x2cos⁡(x3)sin⁡(x3)\dfrac{3x^2\cos(x^3)}{\sin(x^3)}sin(x3)3x2cos(x3)​
  4. 3x2cos⁡(x3)3x^2\cos(x^3)3x2cos(x3)
  5. cos⁡(x3)sin⁡(x3)⋅x2\dfrac{\cos(x^3)}{\sin(x^3)}\cdot x^2sin(x3)cos(x3)​⋅x2
Explanation: This is ln composed with sin(x³). The derivative of ln(u) is 1/u, giving us 1/sin(x³) times the derivative of sin(x³). The derivative of sin(x³) requires another application of the chain rule: cos(x³) · 3x². Combining these: q'(x) = (1/sin(x³)) · cos(x³) · 3x² = 3x²cos(x³)/sin(x³). This can also be written as 3x²cot(x³). A common error is forgetting the 3x² from differentiating x³. Pattern tip: ln(sin(power function)) derivatives always include the power function's derivative as a factor in the numerator.

Question 7

A function is H(x)=cos⁡ ⁣(ln⁡ ⁣(x2+1x))H(x)=\cos\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)H(x)=cos(ln(xx2+1​)). What is H′(x)H'(x)H′(x)?

  1. −sin⁡ ⁣(ln⁡ ⁣(x2+1x))-\sin\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)−sin(ln(xx2+1​))
  2. −sin⁡ ⁣(ln⁡ ⁣(x2+1x))x2+1x-\dfrac{\sin\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)}{\frac{x^2+1}{x}}−xx2+1​sin(ln(xx2+1​))​
  3. −(x2−1x(x2+1))sin⁡ ⁣(ln⁡ ⁣(x2+1x))-\left(\dfrac{x^2-1}{x(x^2+1)}\right)\sin\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)−(x(x2+1)x2−1​)sin(ln(xx2+1​))
  4. (x2−1x(x2+1))sin⁡ ⁣(ln⁡ ⁣(x2+1x))\left(\dfrac{x^2-1}{x(x^2+1)}\right)\sin\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)(x(x2+1)x2−1​)sin(ln(xx2+1​))
  5. −(x2+1x)sin⁡ ⁣(ln⁡ ⁣(x2+1x))-\left(\dfrac{x^2+1}{x}\right)\sin\!\left(\ln\!\left(\dfrac{x^2+1}{x}\right)\right)−(xx2+1​)sin(ln(xx2+1​))
Explanation: Differentiating H(x) = cos(ln((x² + 1)/x)) uses chain. Outer cos(k), derivative -sin(k), k = ln((x² + 1)/x) = ln(x² + 1) - ln x. k' = (2x/(x² + 1)) - 1/x = (x² - 1)/(x (x² + 1)). So H'(x) = -sin(k) * (x² - 1)/(x (x² + 1)). A common omission is wrong sign in k'. Pattern: Simplify log quotients before chaining with outer trig.

Question 8

For f(t)=ln⁡ ⁣(5−2cos⁡(3t))f(t)=\ln\!\left(5-2\cos(3t)\right)f(t)=ln(5−2cos(3t)), what is f′(t)f'(t)f′(t)?

  1. 15−2cos⁡(3t)\dfrac{1}{5-2\cos(3t)}5−2cos(3t)1​
  2. 6sin⁡(3t)5−2cos⁡(3t)\dfrac{6\sin(3t)}{5-2\cos(3t)}5−2cos(3t)6sin(3t)​
  3. 2sin⁡(3t)5−2cos⁡(3t)\dfrac{2\sin(3t)}{5-2\cos(3t)}5−2cos(3t)2sin(3t)​
  4. −6sin⁡(3t)5−2cos⁡(3t)\dfrac{-6\sin(3t)}{5-2\cos(3t)}5−2cos(3t)−6sin(3t)​
  5. 15−2cos⁡(3t)⋅(−2sin⁡(3t))\dfrac{1}{5-2\cos(3t)}\cdot(-2\sin(3t))5−2cos(3t)1​⋅(−2sin(3t))
Explanation: To find the derivative of f(t) = ln(5 - 2cos(3t)), we apply the chain rule with a logarithmic outer function and a trigonometric expression inside. The derivative of ln(u) is 1/u, so we get 1/(5 - 2cos(3t)) as the outer derivative. For the inner function 5 - 2cos(3t), we differentiate to get -2(-sin(3t))·3 = 6sin(3t), where we must remember both the chain rule for cos(3t) and the coefficient -2. Multiplying outer and inner derivatives gives f'(t) = [1/(5 - 2cos(3t))]·6sin(3t) = 6sin(3t)/(5 - 2cos(3t)). A common error is forgetting to multiply by 3 when differentiating cos(3t). When you see ln(trig expression), always identify the outer logarithm and inner trigonometric function, then multiply their derivatives systematically.

Question 9

If u(x)=tan⁡ ⁣(ln⁡(x2+1))u(x)=\tan\!\left(\ln(x^2+1)\right)u(x)=tan(ln(x2+1)), what is u′(x)u'(x)u′(x)?

  1. sec⁡2 ⁣(ln⁡(x2+1))⋅1x2+1\sec^2\!\left(\ln(x^2+1)\right)\cdot\dfrac{1}{x^2+1}sec2(ln(x2+1))⋅x2+11​
  2. sec⁡2 ⁣(ln⁡(x2+1))⋅2xx2+1\sec^2\!\left(\ln(x^2+1)\right)\cdot\dfrac{2x}{x^2+1}sec2(ln(x2+1))⋅x2+12x​
  3. sec⁡2 ⁣(ln⁡(x2+1))\sec^2\!\left(\ln(x^2+1)\right)sec2(ln(x2+1))
  4. tan⁡ ⁣(ln⁡(x2+1))⋅2xx2+1\tan\!\left(\ln(x^2+1)\right)\cdot\dfrac{2x}{x^2+1}tan(ln(x2+1))⋅x2+12x​
  5. 2xx2+1\dfrac{2x}{x^2+1}x2+12x​
Explanation: We have tan composed with ln(x²+1). The derivative of tan(u) is sec²(u), so we get sec²(ln(x²+1)) times the derivative of ln(x²+1). The derivative of ln(x²+1) is 1/(x²+1) times 2x (from differentiating x²+1). Therefore, u'(x) = sec²(ln(x²+1)) · 2x/(x²+1). A common error is forgetting the 2x from the derivative of x²+1 or confusing tan's derivative with sin or cos derivatives. Remember: tan(ln(polynomial)) produces sec² of the same argument, always with the polynomial's derivative in the numerator of the accompanying fraction.

Question 10

If s(x)=sec⁡ ⁣((x−2)5)s(x)=\sec\!\left((x-2)^5\right)s(x)=sec((x−2)5), what is s′(x)s'(x)s′(x)?

  1. sec⁡ ⁣((x−2)5)tan⁡ ⁣((x−2)5)\sec\!\left((x-2)^5\right)\tan\!\left((x-2)^5\right)sec((x−2)5)tan((x−2)5)
  2. 5(x−2)4sec⁡ ⁣((x−2)5)tan⁡ ⁣((x−2)5)5(x-2)^4\sec\!\left((x-2)^5\right)\tan\!\left((x-2)^5\right)5(x−2)4sec((x−2)5)tan((x−2)5)
  3. sec⁡ ⁣((x−2)5)tan⁡ ⁣((x−2)5)⋅(x−2)4\sec\!\left((x-2)^5\right)\tan\!\left((x-2)^5\right)\cdot(x-2)^4sec((x−2)5)tan((x−2)5)⋅(x−2)4
  4. 5(x−2)4sec⁡2 ⁣((x−2)5)5(x-2)^4\sec^2\!\left((x-2)^5\right)5(x−2)4sec2((x−2)5)
  5. tan⁡ ⁣((x−2)5)⋅5(x−2)4\tan\!\left((x-2)^5\right)\cdot 5(x-2)^4tan((x−2)5)⋅5(x−2)4
Explanation: This involves sec composed with (x-2)⁵. The derivative of sec(u) is sec(u)tan(u), and we multiply by the derivative of (x-2)⁵. Using the chain rule: d/dx[(x-2)⁵] = 5(x-2)⁴. Therefore, s'(x) = sec((x-2)⁵)tan((x-2)⁵) · 5(x-2)⁴. A common mistake is confusing the derivative of sec with that of tan (which would give sec²). Remember the mnemonic: the derivative of sec involves both sec and tan. Pattern: sec(power function) derivatives always include the original sec, tan of the same argument, and the power rule applied to the inner function.

Question 11

Let N(x)=tan⁡ ⁣(2x2−1)N(x)=\tan\!\left(\dfrac{2}{x^2-1}\right)N(x)=tan(x2−12​). What is N′(x)N'(x)N′(x)?

  1. sec⁡2 ⁣(2x2−1)\sec^2\!\left(\dfrac{2}{x^2-1}\right)sec2(x2−12​)
  2. −4x(x2−1)2sec⁡2 ⁣(2x2−1)-\dfrac{4x}{(x^2-1)^2}\sec^2\!\left(\dfrac{2}{x^2-1}\right)−(x2−1)24x​sec2(x2−12​)
  3. 4x(x2−1)2sec⁡2 ⁣(2x2−1)\dfrac{4x}{(x^2-1)^2}\sec^2\!\left(\dfrac{2}{x^2-1}\right)(x2−1)24x​sec2(x2−12​)
  4. −4(x2−1)2sec⁡2 ⁣(2x2−1)-\dfrac{4}{(x^2-1)^2}\sec^2\!\left(\dfrac{2}{x^2-1}\right)−(x2−1)24​sec2(x2−12​)
  5. −4x(x2−1)2tan⁡ ⁣(2x2−1)-\dfrac{4x}{(x^2-1)^2}\tan\!\left(\dfrac{2}{x^2-1}\right)−(x2−1)24x​tan(x2−12​)
Explanation: For N(x) = tan(2/(x² - 1)), chain rule. Outer tan(r), derivative sec²(r), r = 2 (x² - 1)^{-1}. r' = 2 * (-1) (x² - 1)^{-2} * 2x = -4x / (x² - 1)². Thus, N'(x) = sec²(r) * (-4x / (x² - 1)²). A common omission is coefficient error in r'. Recognize trig of rational with quadratic denominator and use power chain.

Question 12

A sensor records f(x)=sin⁡ ⁣((3x−2)5)f(x)=\sin\!\left((3x-2)^5\right)f(x)=sin((3x−2)5). What is f′(x)f'(x)f′(x)?

  1. cos⁡ ⁣((3x−2)5)\cos\!\left((3x-2)^5\right)cos((3x−2)5)
  2. 15(3x−2)4cos⁡ ⁣((3x−2)5)15(3x-2)^4\cos\!\left((3x-2)^5\right)15(3x−2)4cos((3x−2)5)
  3. 5(3x−2)4cos⁡ ⁣((3x−2)5)5(3x-2)^4\cos\!\left((3x-2)^5\right)5(3x−2)4cos((3x−2)5)
  4. sin⁡ ⁣((3x−2)5)⋅15(3x−2)4\sin\!\left((3x-2)^5\right)\cdot 15(3x-2)^4sin((3x−2)5)⋅15(3x−2)4
  5. 3cos⁡ ⁣((3x−2)5)3\cos\!\left((3x-2)^5\right)3cos((3x−2)5)
Explanation: This problem requires finding the derivative of a composite function involving a trigonometric sine function with a polynomial inner expression, necessitating the chain rule. The outer function is sin(u), where u = (3x - 2)^5, and its derivative is cos(u), while the inner function u has derivative 5(3x - 2)^4 * 3 = 15(3x - 2)^4. To find f'(x), multiply cos((3x - 2)^5) by 15(3x - 2)^4, yielding the full derivative. A common omission is forgetting to apply the chain rule to the power function inside, such as missing the factor of 5 from the exponent or the 3 from the linear term. Another frequent mistake is confusing the derivative of sine with cosine without the negative sign, but here it's positive for sine. By recognizing the nested structure—trig function of a power of a linear expression—you can apply the chain rule in layers, starting from the outermost. This pattern of multiple compositions is common in calculus problems, so practice decomposing functions into outer and inner parts to build proficiency.

Question 13

Let r(x)=ln⁡ ⁣(sin⁡xx)r(x)=\ln\!\left(\dfrac{\sin x}{x}\right)r(x)=ln(xsinx​). What is r′(x)r'(x)r′(x)?

  1. cos⁡xsin⁡x\dfrac{\cos x}{\sin x}sinxcosx​
  2. cos⁡xsin⁡x−1x\dfrac{\cos x}{\sin x}-\dfrac{1}{x}sinxcosx​−x1​
  3. cos⁡xx−sin⁡xx2\dfrac{\cos x}{x}-\dfrac{\sin x}{x^2}xcosx​−x2sinx​
  4. 1sin⁡x/x\dfrac{1}{\sin x/x}sinx/x1​
  5. ln⁡ ⁣(sin⁡xx)(cos⁡xx−sin⁡xx2)\ln\!\left(\dfrac{\sin x}{x}\right)\left(\dfrac{\cos x}{x}-\dfrac{\sin x}{x^2}\right)ln(xsinx​)(xcosx​−x2sinx​)
Explanation: For r(x) = ln(sin x / x), recognize this as a logarithmic function of a quotient, which can be rewritten as ln(sin x) - ln x for easier differentiation. The derivative involves the chain rule on each term: for ln(sin x), it's (1/sin x) * cos x = cot x, and for -ln x, it's -1/x. Combining gives cot x - 1/x, or (cos x / sin x) - 1/x. The outer log function's derivative is 1/u times u', where u = sin x / x. A common omission is failing to apply the quotient rule to u or mishandling the log property, resulting in incorrect terms. To spot these patterns, look for logs of ratios or products and simplify using log rules before differentiating with the chain rule.

Question 14

For m(x)=ln⁡ ⁣(1+sin⁡x)m(x)=\ln\!\left(\sqrt{1+\sin x}\right)m(x)=ln(1+sinx​), what is m′(x)m'(x)m′(x)?

  1. cos⁡x2(1+sin⁡x)\dfrac{\cos x}{2(1+\sin x)}2(1+sinx)cosx​
  2. 11+sin⁡x\dfrac{1}{\sqrt{1+\sin x}}1+sinx​1​
  3. cos⁡x1+sin⁡x\dfrac{\cos x}{\sqrt{1+\sin x}}1+sinx​cosx​
  4. 121+sin⁡x\dfrac{1}{2\sqrt{1+\sin x}}21+sinx​1​
  5. cos⁡x21+sin⁡x\dfrac{\cos x}{2\sqrt{1+\sin x}}21+sinx​cosx​
Explanation: We can simplify first: ln(√(1+sin x)) = ln((1+sin x)^(1/2)) = (1/2)ln(1+sin x). Now differentiating: m'(x) = (1/2) · 1/(1+sin x) · cos x = cos x/(2(1+sin x)). Alternatively, treating it as a composition: derivative of ln(u) is 1/u, so 1/√(1+sin x), times the derivative of √(1+sin x), which is cos x/(2√(1+sin x)). This gives cos x/(2(1+sin x)) after simplification. Students often forget to apply logarithm properties or make errors with the square root derivative. Pattern tip: ln(√(expression)) can be simplified using log properties before differentiating, often making the work easier.

Question 15

A signal is modeled by p(x)=cos⁡ ⁣(ln⁡(2x+1))p(x)=\cos\!\left(\ln(2x+1)\right)p(x)=cos(ln(2x+1)). What is p′(x)p'(x)p′(x)?

  1. −sin⁡ ⁣(ln⁡(2x+1))-\sin\!\left(\ln(2x+1)\right)−sin(ln(2x+1))
  2. −sin⁡ ⁣(ln⁡(2x+1))2x+1-\dfrac{\sin\!\left(\ln(2x+1)\right)}{2x+1}−2x+1sin(ln(2x+1))​
  3. −2sin⁡ ⁣(ln⁡(2x+1))2x+1-\dfrac{2\sin\!\left(\ln(2x+1)\right)}{2x+1}−2x+12sin(ln(2x+1))​
  4. 2cos⁡ ⁣(ln⁡(2x+1))2x+1\dfrac{2\cos\!\left(\ln(2x+1)\right)}{2x+1}2x+12cos(ln(2x+1))​
  5. −2sin⁡ ⁣(ln⁡(2x+1))-2\sin\!\left(\ln(2x+1)\right)−2sin(ln(2x+1))
Explanation: Here, we differentiate a cosine function composed with a logarithm, invoking the chain rule. Outer is cos(u), u = ln(2x + 1), derivative -sin(u), inner u' = 2/(2x + 1). So p'(x) = -sin(ln(2x + 1)) * 2/(2x + 1). Common omission: forgetting the factor of 2 from the inner log's derivative. Another mistake is omitting the negative sign from cosine's derivative. Identify the trig of log pattern and chain accordingly. This recognition transfers to nested functions, improving problem-solving.

Question 16

A function is s(x)=cos⁡ ⁣(2x2−3x)s(x)=\cos\!\left(\sqrt{2x^2-3x}\right)s(x)=cos(2x2−3x​). What is s′(x)s'(x)s′(x)?

  1. −sin⁡ ⁣(2x2−3x)-\sin\!\left(\sqrt{2x^2-3x}\right)−sin(2x2−3x​)
  2. −(4x−3)sin⁡ ⁣(2x2−3x)2x2−3x-\dfrac{(4x-3)\sin\!\left(\sqrt{2x^2-3x}\right)}{\sqrt{2x^2-3x}}−2x2−3x​(4x−3)sin(2x2−3x​)​
  3. −(4x−3)sin⁡ ⁣(2x2−3x)22x2−3x-\dfrac{(4x-3)\sin\!\left(\sqrt{2x^2-3x}\right)}{2\sqrt{2x^2-3x}}−22x2−3x​(4x−3)sin(2x2−3x​)​
  4. −(4x−3)cos⁡ ⁣(2x2−3x)22x2−3x-\dfrac{(4x-3)\cos\!\left(\sqrt{2x^2-3x}\right)}{2\sqrt{2x^2-3x}}−22x2−3x​(4x−3)cos(2x2−3x​)​
  5. −(2x2−3x)sin⁡ ⁣(2x2−3x)22x2−3x-\dfrac{(2x^2-3x)\sin\!\left(\sqrt{2x^2-3x}\right)}{2\sqrt{2x^2-3x}}−22x2−3x​(2x2−3x)sin(2x2−3x​)​
Explanation: To differentiate s(x) = cos(√(2x² - 3x)), note the multiple layers requiring repeated chain rule application. The outermost is cos(v), derivative -sin(v), where v = √(2x² - 3x). Then v's derivative is (1/(2√(2x² - 3x))) * (4x - 3). Combining yields -sin(v) * (4x - 3)/(2√(2x² - 3x)). A common omission is forgetting the 1/2 in the square root derivative, leading to an extra factor of 2 in the denominator. For pattern recognition, identify multi-layer compositions like trig of radical of polynomial and apply chain rule stepwise from outside in.

Question 17

A function is p(x)=tan⁡ ⁣(x2+13)p(x)=\tan\!\left(\dfrac{x^2+1}{3}\right)p(x)=tan(3x2+1​). What is p′(x)p'(x)p′(x)?

  1. sec⁡2 ⁣(x2+13)\sec^2\!\left(\dfrac{x^2+1}{3}\right)sec2(3x2+1​)
  2. 2x3sec⁡2 ⁣(x2+13)\dfrac{2x}{3}\sec^2\!\left(\dfrac{x^2+1}{3}\right)32x​sec2(3x2+1​)
  3. 2x3tan⁡ ⁣(x2+13)\dfrac{2x}{3}\tan\!\left(\dfrac{x^2+1}{3}\right)32x​tan(3x2+1​)
  4. 2xsec⁡2 ⁣(x2+13)2x\sec^2\!\left(\dfrac{x^2+1}{3}\right)2xsec2(3x2+1​)
  5. 13sec⁡2 ⁣(x2+13)\dfrac{1}{3}\sec^2\!\left(\dfrac{x^2+1}{3}\right)31​sec2(3x2+1​)
Explanation: To find the derivative of p(x) = tan((x² + 1)/3), we apply the chain rule because this is a trigonometric function composed with a polynomial expression. The outer function is tan(u), where u = (x² + 1)/3, and its derivative is sec²(u). The inner function u has derivative (2x)/3. Therefore, p'(x) = sec²((x² + 1)/3) * (2x)/3, which simplifies to the given expression. A common omission is forgetting to multiply by the derivative of the inner function, leading to answers like sec²((x² + 1)/3) without the (2x)/3 factor. To recognize such patterns in the future, identify nested functions where a trig function encloses a non-constant expression and apply the chain rule by differentiating outer first then multiplying by inner's derivative.

Question 18

Let J(x)=ln⁡ ⁣(ln⁡(x2+e))J(x)=\ln\!\left(\ln(x^2+e)\right)J(x)=ln(ln(x2+e)). What is J′(x)J'(x)J′(x)?

  1. 1ln⁡(x2+e)\dfrac{1}{\ln(x^2+e)}ln(x2+e)1​
  2. 2xx2+e\dfrac{2x}{x^2+e}x2+e2x​
  3. 2xln⁡(x2+e)\dfrac{2x}{\ln(x^2+e)}ln(x2+e)2x​
  4. 2x(x2+e)ln⁡(x2+e)\dfrac{2x}{(x^2+e)\ln(x^2+e)}(x2+e)ln(x2+e)2x​
  5. ln⁡(ln⁡(x2+e))⋅2xx2+e\ln(\ln(x^2+e))\cdot \dfrac{2x}{x^2+e}ln(ln(x2+e))⋅x2+e2x​
Explanation: For J(x) = ln(ln(x² + e)), double log requires repeated chain. Outer ln(m), derivative 1/m, m = ln(x² + e). m' = (1/(x² + e)) * 2x. Thus, J'(x) = (1/m) * (2x/(x² + e)). A common omission is forgetting the outer 1/m. Recognize nested logs and apply chain sequentially.

Question 19

A function is B(x)=tan⁡ ⁣(ln⁡ ⁣(1x))B(x)=\tan\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)B(x)=tan(ln(x1​)). What is B′(x)B'(x)B′(x)?

  1. sec⁡2 ⁣(ln⁡ ⁣(1x))\sec^2\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)sec2(ln(x1​))
  2. −1xsec⁡2 ⁣(ln⁡ ⁣(1x))-\dfrac{1}{x}\sec^2\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)−x1​sec2(ln(x1​))
  3. 1xsec⁡2 ⁣(ln⁡ ⁣(1x))\dfrac{1}{x}\sec^2\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)x1​sec2(ln(x1​))
  4. −1x2sec⁡2 ⁣(ln⁡ ⁣(1x))-\dfrac{1}{x^2}\sec^2\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)−x21​sec2(ln(x1​))
  5. −1xtan⁡ ⁣(ln⁡ ⁣(1x))-\dfrac{1}{x}\tan\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)−x1​tan(ln(x1​))
Explanation: To differentiate B(x) = tan(ln(1/x)), apply chain rule. Outer tan(e), derivative sec²(e), e = ln(1/x) = -ln x. e' = -1/x. So B'(x) = sec²(e) * (-1/x). A common omission is forgetting the negative from the inner log's derivative. Pattern strategy: Note trig of log of reciprocal and simplify log first to capture the sign in the chain.

Question 20

Let Q(x)=ln⁡ ⁣(1+sin⁡ ⁣(x2))Q(x)=\ln\!\left(1+\sin\!\left(x^2\right)\right)Q(x)=ln(1+sin(x2)). What is Q′(x)Q'(x)Q′(x)?

  1. 11+sin⁡(x2)\dfrac{1}{1+\sin(x^2)}1+sin(x2)1​
  2. 2xcos⁡(x2)1+sin⁡(x2)\dfrac{2x\cos(x^2)}{1+\sin(x^2)}1+sin(x2)2xcos(x2)​
  3. cos⁡(x2)1+sin⁡(x2)\dfrac{\cos(x^2)}{1+\sin(x^2)}1+sin(x2)cos(x2)​
  4. 2xcos⁡(x2)2x\cos(x^2)2xcos(x2)
  5. ln⁡(1+sin⁡(x2))⋅2xcos⁡(x2)\ln(1+\sin(x^2))\cdot 2x\cos(x^2)ln(1+sin(x2))⋅2xcos(x2)
Explanation: For Q(x) = ln(1 + sin(x²)), chain rule. Outer ln(t), derivative 1/t, t = 1 + sin(x²). t' = cos(x²) * 2x. Thus, Q'(x) = (1/(1 + sin(x²))) * 2x cos(x²). A common omission is missing 2x. Recognize logs of summed trig with inner quadratic and chain accordingly.