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).

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.

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.

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.

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.

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 $\frac{1}{|u| \sqrt{u^2 - 1}} \times u'$, which is 3. Thus, $\frac{3}{|3x-2| \sqrt{(3x-2)^2 - 1}}$. 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.

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.

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.

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'.

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.