This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsec function. Derivative of arcsec(u) with u as 2-x is 1 over |u| sqrt(u squared minus 1) times -1. Gives -1 over |2-x| sqrt(...). Negative from decreasing u. Choice A uses positive, ignoring u' sign. Pattern: Decreasing arguments in arcsec introduce a negative sign via chain rule.

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.

This problem involves differentiating arctangent of a linear expression. The derivative of arctan(u) is 1/(1+u²) times the derivative of u. Here, u = 1-4x, so we multiply by -4 (the derivative of 1-4x). This yields -4/(1+(1-4x)²), keeping the entire expression (1-4x) squared in the denominator. A student might incorrectly write 4/(1+(1-4x)²) by missing the negative sign from the derivative of 1-4x. When the inner function decreases (has a negative derivative), the overall derivative must reflect this negative contribution.

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.

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'.

This problem tests the skill of differentiating inverse trigonometric functions, specifically the arccos function. The derivative of arccos(u) with u as 7-2x is -1 over sqrt(1 minus u squared) times u', where u' is -2. Multiplying yields positive 2 over the sqrt term due to the negatives canceling. This illustrates how the chain rule can flip signs based on the inner derivative. Choice B forgets to account for the positive outcome from the chain rule. Recognize patterns in arccos with decreasing inners by checking if signs cancel to positive.

This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. For arctan(u) with u as 1-3x, the derivative is 1 over 1 plus u squared times u', which is -3. This results in -3 over 1 plus (1-3x) squared, incorporating the chain rule. The negative reflects the decreasing inner function. Choice A ignores the negative from u', leading to a sign error. Spot arctan patterns with negative coefficients by ensuring the multiplier carries the sign.

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.

This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsin function. Derivative of arcsin(u) with u as x/4 is 1 over sqrt(1 minus u squared) times 1/4. This yields 1/4 over sqrt(1 minus (x/4) squared), adjusting for the scaled input. It shows how fractions in arguments reduce the derivative's magnitude. Choice A forgets the 1/4 chain rule factor, overestimating the rate. Recognize scaled arcsin by dividing the standard derivative by the denominator in u.

This problem tests the skill of differentiating inverse trigonometric functions, specifically the arcsin function. The derivative of arcsin(u) is 1 over the square root of 1 minus u squared, multiplied by the derivative of u. Here, u is 3x, so u' is 3, and we multiply the standard derivative by 3 to account for the chain rule. This results in 3 divided by the square root of 1 minus 9x squared, capturing how the rate of change scales with the coefficient. A tempting distractor is choice B, which omits the chain rule multiplier of 3 and underestimates the derivative. To recognize similar patterns, look for arcsin of a linear term and always multiply by the inner function's derivative.