This question tests the derivative of the cotangent function, another reciprocal trigonometric function. The derivative of cot(t) is -csc²(t), derived from cot(t) = cos(t)/sin(t) using the quotient rule. For f(t) = cot(t) - 5t, we apply this to get f'(t) = -csc²(t) - 5. The linear term -5t contributes -5 to the derivative. Option A with positive csc²(t) - 5 is incorrect because it misses the negative sign in the cotangent derivative. Remember that both cot and tan have negative derivatives involving squared functions.

This question assesses differentiating reciprocal trig functions, particularly cotangent with a fractional constant. The derivative of cot(t) is -csc²(t), and multiplying by 1/2 yields -1/2 csc²(t). This stems from the quotient rule derivation of cotangent. It's applicable in control functions. One might pick 1/2 csc²(t), but it fails by missing the negative sign. A transferable approach is to derive unfamiliar trig derivatives from basic sine and cosine if needed.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The derivative of sec(t) is sec(t) tan(t), as it follows from the quotient rule applied to sec(t) = 1/cos(t). When multiplied by a constant like 4, the derivative becomes 4 sec(t) tan(t) due to the constant multiple rule. This formula is essential for functions involving secant in modeling scenarios like height over time. A tempting distractor might be -4 sec(t) tan(t), but it fails because the derivative of secant is positive sec tan, not negative. Always remember to recall the standard derivatives of trig functions and apply sign rules carefully when constants are involved.

This problem requires differentiating reciprocal trigonometric functions, specifically the cotangent function. The derivative of cot(u) is -csc²(u)·u', where u' is the derivative of the inner function. For T(x) = 7cot(3x), we apply the chain rule: T'(x) = 7·(-csc²(3x))·3 = -21csc²(3x). A common mistake would be to use the cosecant derivative formula -csc(u)cot(u) instead of -csc²(u), which would give choice E: -21csc(3x)cot(3x). When differentiating reciprocal trig functions, match each function with its correct derivative: cot has csc², while csc has csc·cot.

This problem involves differentiating reciprocal trigonometric functions, specifically the cosecant function. The derivative of csc(θ) is -csc(θ)cot(θ), and the derivative of θ² is 2θ. For h(θ) = 4csc(θ) + θ², we get h'(θ) = 4·(-csc(θ)cot(θ)) + 2θ = -4csc(θ)cot(θ) + 2θ. A tempting error would be to use the secant derivative formula instead, giving -4sec(θ)tan(θ) + 2θ, which appears as choice C. Remember that csc(θ) = 1/sin(θ) has a negative sign in its derivative, while sec(θ) = 1/cos(θ) has a positive sign in its derivative.

This problem requires differentiating reciprocal trigonometric functions, specifically the secant function. The derivative of sec(u) is sec(u)tan(u)·u', where u' is the derivative of the inner function. For s(t) = 5sec(2t) - 3, we apply the chain rule: s'(t) = 5·sec(2t)tan(2t)·2 - 0 = 10sec(2t)tan(2t). A common error would be forgetting the chain rule factor of 2 from differentiating the inner function 2t, which would give 5sec(2t)tan(2t) instead. When differentiating reciprocal trig functions, always remember to multiply by the derivative of the inner function and use the correct derivative formula for each reciprocal function.

This question combines the cotangent derivative with chain rule and polynomial differentiation. The derivative of cot(u) is -csc²(u), and for u = 5x, we multiply by 5. For C(x) = 8cot(5x) - x², we get C'(x) = 8·(-csc²(5x))·5 - 2x = -40csc²(5x) - 2x. The factors 8, -1, and 5 multiply to give -40, while -x² contributes -2x. Option A with positive 40csc²(5x) - 2x incorrectly omits the negative from the cotangent derivative. When differentiating expressions with multiple terms, handle each term's derivative separately then combine.

This question tests knowledge of differentiating the cosecant function, a reciprocal trigonometric function. The derivative of csc(x) is -csc(x)cot(x), which can be derived from csc(x) = 1/sin(x) using the quotient rule. For h(x) = csc(x) + 3x, we get h'(x) = -csc(x)cot(x) + 3. Option D might tempt students who forget the negative sign in the derivative of csc(x), but the derivative must be negative. To remember reciprocal trig derivatives, note that derivatives of co-functions (cosecant, cotangent) include negative signs.

This question involves the product rule combined with differentiating the cotangent function, a reciprocal trigonometric function. The derivative of cot(t) is -csc²(t), and we apply the product rule: (uv)' = u'v + uv'. For g(t) = t·cot(t), we get g'(t) = 1·cot(t) + t·(-csc²(t)) = cot(t) - t·csc²(t). Option B incorrectly has a positive sign for the second term, missing the negative in cot's derivative. When combining product rule with reciprocal trig derivatives, carefully track all negative signs from the derivative formulas.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The derivative of sec(t) is sec(t) tan(t), derived using the quotient rule on 1/cos(t), which gives (sin(t)/cos²(t)) or sec(t) tan(t). When multiplied by the constant 5, the derivative becomes 5 sec(t) tan(t). For the given domain 0 < t < π/2, this positive form holds as both sec and tan are positive there. A tempting distractor like choice C fails because it includes an unnecessary negative sign, which would apply to csc(t) instead. Always remember to apply the chain rule when the argument is more complex than just the variable, though here it's straightforward.