This problem involves differentiating the cosecant function with a linear argument. The derivative of csc(u) is -csc(u)cot(u) times the derivative of u. For s(x) = csc(4x), the chain rule gives s'(x) = -csc(4x)cot(4x) · 4 = -4csc(4x)cot(4x). A typical error is forgetting the negative sign inherent in cosecant's derivative (choice A), which stems from not memorizing the formula correctly. The strategy for success is to remember that csc and cot derivatives always carry negative signs, unlike their reciprocal counterparts sec and tan.

This question tests differentiation of the cotangent function with a composite argument. The derivative of cot(u) is -csc²(u) times the derivative of u. For g(θ) = cot(5θ), applying the chain rule gives g'(θ) = -csc²(5θ) · 5 = -5csc²(5θ). A common error is confusing cotangent's derivative with tangent's derivative, leading to sec² instead of csc² (choice D). Remember that cot and csc are paired in derivatives, just as tan and sec are paired, and both cot and csc derivatives include negative signs.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is q(x) = csc(x - 2), so let u = x - 2 with u' = 1. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', resulting in -csc(x - 2) cot(x - 2). This shows the negative sign's importance. A tempting distractor is choice A, which forgets the negative sign inherent to cosecant. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is C(t) = csc(2 - t), so let u = 2 - t with u' = -1. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', yielding csc(2 - t) cot(2 - t) from double negatives. This handles the decreasing linear inner. A tempting distractor is choice A, which retains a negative by sign error. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. D(x) = sec(sin x), derivative is sec(u) tan(u) u' where u = sin x. u' = cos x. Thus, D'(x) = sec(sin x) tan(sin x) * cos x. Choice A misses the cos x factor, tempting without chain rule, but composition requires it. When trig functions are nested, apply chain rule multiple times if necessary for accurate derivatives.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is P(t) = sec(t - π/6) + 1, so the constant 1 differentiates to 0. Let u = t - π/6 with u' = 1, so the derivative is sec(u) tan(u) multiplied by 1, or sec(t - π/6) tan(t - π/6). This ignores the added constant. A tempting distractor is choice C, which confuses with tangent's derivative. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. φ(x) = csc(π x²), derivative -csc(u) cot(u) u' with u = π x². u' = 2 π x. So -csc(π x²) cot(π x²) * 2 π x = -2 π x csc(π x²) cot(π x²). Choice C forgets the 2x factor, incomplete chain rule. Include all parts of the inner derivative, like product rule if needed, for composites.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cotangent function. The function is $f(t) = \cot(5t)$, so let $u = 5t$ with $u' = 5$. The derivative of $\cot(u)$ is $-\csc^2(u)$ multiplied by $u'$, resulting in $-5 \csc^2(5t)$. This structure emphasizes the negative sign and the squared cosecant term. A tempting distractor is choice E, which confuses the cotangent derivative with that of cosecant, leading to an incorrect form. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is u(x) = sec(√x), so let v = √x with v' = 1/(2√x). The derivative of sec(v) is sec(v) tan(v) multiplied by v', yielding (1/(2√x)) sec(√x) tan(√x). This incorporates the square root derivative. A tempting distractor is choice A, which forgets the 1/(2√x) chain factor. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is I(t) = sec(t), with no inner composition beyond t itself. The derivative of sec(t) is sec(t) tan(t), as the chain multiplier is 1. This basic form recalls the fundamental secant derivative. A tempting distractor is choice B, which confuses it with the tangent derivative of sec²(t). When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.