This problem tests the skill of differentiating reciprocal trigonometric functions, recognizing 1/tan(x) as cot(x). The derivative of cot(x) is -csc²(x), derived from the quotient rule on cos(x)/sin(x). This negative form distinguishes it from tan(x)'s derivative. For 0 < x < π, the function is defined appropriately. A tempting distractor like choice A fails by using csc cot, which is actually related to csc's derivative without the negative. Always remember to apply the chain rule when the argument is more complex than just the variable, though here it's straightforward.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically cotangent with a chain rule. The derivative of cot(5x) is -csc²(5x) times 5, resulting in -5 csc²(5x). This applies the chain rule to the standard cot derivative of -csc²(u) where u=5x. The negative sign is key to cotangent differentiation. A tempting distractor like choice E fails by using -csc cot instead of -csc², confusing it with csc's derivative. Always remember to apply the chain rule when the argument is more complex than just the variable, multiplying by the inner derivative.

This problem requires differentiating a reciprocal trigonometric function, specifically the secant function. The derivative of sec(t) is sec(t)tan(t), which comes from the fact that sec(t) = 1/cos(t) and using the chain rule. For s(t) = 7sec(t) - 3, we apply the constant multiple rule to get s'(t) = 7·sec(t)tan(t) - 0 = 7sec(t)tan(t). The constant term -3 has derivative 0. A common error would be to confuse this with the cosecant derivative formula, which would give option D with -7csc(t)cot(t). Remember that sec and tan go together in derivatives, just as csc and cot do.

This problem combines the secant derivative with the chain rule for composite functions. The derivative of sec(u) is sec(u)tan(u), and when u = 3θ, we must multiply by the derivative of the inner function. For h(θ) = 2sec(3θ), we get h'(θ) = 2·sec(3θ)tan(3θ)·3 = 6sec(3θ)tan(3θ). The chain rule contributes the factor of 3 from the derivative of 3θ. Option B with only 2sec(3θ)tan(3θ) forgets to apply the chain rule. When differentiating reciprocal trig functions with composite arguments, always multiply by the derivative of the inner function.

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 requires differentiating cosecant with a composite argument using the chain rule. The derivative of csc(u) is -csc(u)cot(u), and for u = 2x, we multiply by 2. For I(x) = 5csc(2x), we get I'(x) = 5·(-csc(2x)cot(2x))·2 = -10csc(2x)cot(2x). The coefficient 5, the negative from the csc derivative, and the chain rule factor 2 combine to give -10. Option A with positive 10csc(2x)cot(2x) misses the crucial negative sign in the cosecant derivative formula. Always remember that csc and cot derivatives carry negative signs.

This problem requires differentiating reciprocal trigonometric functions, specifically the cosecant function within a linear combination. The derivative of -csc(u) is csc(u)cot(u)·u' (note the sign change), and for r(x) = 2x - 9csc(5x), we get r'(x) = 2 - 9·(-csc(5x)cot(5x))·5 = 2 + 45csc(5x)cot(5x). A tempting mistake would be to use the secant derivative formula, giving 2 + 45sec(5x)tan(5x) as in choice C. Remember that the negative sign in front of csc creates a positive derivative term, and always apply the chain rule for composite arguments.

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 involves the skill of differentiating reciprocal trig functions, focusing on cosecant plus a constant. The derivative of csc(x) is -csc(x) cot(x), so 5 times that is -5 csc(x) cot(x), with +1 differentiating to zero. The sum rule ensures only the trig term contributes. Tracking models use such functions. Choosing 5 csc(x) cot(x) fails by forgetting the negative sign in csc's derivative. To build proficiency, differentiate using quotient rule for reinforcement.