This problem tests the basic derivative rules for constants, sums, and differences. To find s'(x) for $s(x) = -5x^3 + 4x - 9$, apply the power rule to the first term: the derivative of $-5x^3$ is $-53x^{2} = -15x^2$. For the second term, the derivative of $+4x$ is $41x^{0} = 4$. The constant term $-9$ has a derivative of 0, so it is omitted from the result. A tempting distractor is choice B, which subtracts the constant $-9$ again, but constants do not persist in derivatives. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

This problem tests the basic derivative rules for constants, sums, and differences. To find h'(t) for h(t) = 9 - $2t^5$ + $6t^2$, apply the power rule to the second term: the derivative of $-2t^5$ is $-25t^{4}$ = $-10t^4$. For the third term, the derivative of $+6t^2$ is $62t^{1}$ = 12t. The constant term 9 has a derivative of 0, so it is not included. A tempting distractor is choice C, which adds +9, but constants disappear upon differentiation. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

This problem requires finding the marginal revenue using basic derivative rules. We differentiate each term separately: the derivative of 3x⁶ is 18x⁵ (3·6·x⁵), the derivative of -11x³ is -33x² (-11·3·x²), and the derivative of 4x is 4. Combining these gives R'(x) = 18x⁵ - 33x² + 4. Choice B incorrectly writes the last term as 4x instead of 4, failing to recognize that d/dx(4x) = 4, not 4x. The systematic approach is to apply the power rule to each term, remembering that linear terms become constants.

This problem requires finding f'(x) using basic derivative rules on a polynomial function. We differentiate each term: the derivative of -9x⁴ is -36x³ (multiply -9 by 4 and reduce the exponent), the derivative of 7x³ is 21x² (7·3·x²), and the derivative of the constant -5 is 0. Thus, f'(x) = -36x³ + 21x² + 0 = -36x³ + 21x². Choice B incorrectly includes the constant -5 in the derivative, but constants always have zero derivative. The key is to apply the power rule term by term, remembering that constant terms disappear.

This problem assesses your understanding of the basic derivative rules for constants, sums, and differences in polynomials. To compute Q'(t), differentiate each term separately using the power rule. The derivative of $-t^4$ is -1 times $4t^{3}$, which is $-4t^3$. The derivative of $+8t^3$ is 8 times $3t^{2}$, which is $+24t^2$, and the derivative of -16 is 0, so they combine to $-4t^3$ + $24t^2$. A tempting distractor like choice D is $-t^4$ + $8t^3$, but this fails because it doesn't apply the power rule at all. Always remember to apply the power rule to each term independently and set the derivative of constants to zero.

This problem involves finding the marginal profit by differentiating the profit function. Starting with R(x) = -2x⁴ + 13x - 20, we apply the power rule to each term. The derivative of -2x⁴ is -2·4x³ = -8x³, the derivative of 13x is 13·1 = 13, and the derivative of the constant -20 is 0. Thus, R'(x) = -8x³ + 13 + 0 = -8x³ + 13. A mistake would be to write 13x instead of 13 (choice B), forgetting that the derivative of x¹ is 1, not x. Remember that when differentiating x to the first power, the result is simply the coefficient.

This problem tests the basic derivative rules for constants, sums, and differences. To find y'(x) for y(x) = $(3/5)x^5$ - $10x^2$ + 8, apply the power rule to the first term: the derivative of $(3/5)x^5$ is $(3/5)5x^{4}$ = $3x^4$. For the second term, the derivative of $-10x^2$ is $-102x^{1}$ = -20x. The constant term +8 has a derivative of 0, so it is not present. A tempting distractor is choice D, which includes +8, but constants are differentiated to zero. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

This temperature model requires differentiating terms including one in parentheses. Given T(t) = 6t² - (t⁶) + 15t, we differentiate each term: 6t² becomes 6·2t = 12t; -(t⁶) becomes -6t⁵; and 15t becomes 15·1 = 15. Thus, T'(t) = 12t - 6t⁵ + 15. Choice B omits the constant term 15, incorrectly thinking the derivative of 15t is 0, but the derivative of 15t is 15. The parentheses around t⁶ don't change the differentiation—we still apply the power rule normally to get -6t⁵.

This problem requires applying basic derivative rules to find the rate of change of water volume. To find V'(t) from V(t) = 7t⁴ - 3t² + 12, we differentiate each term separately using the power rule. For 7t⁴, we get 7·4t³ = 28t³; for -3t², we get -3·2t = -6t; and the constant 12 becomes 0. Therefore, V'(t) = 28t³ - 6t + 0 = 28t³ - 6t. A common error would be keeping the constant term, giving 28t³ - 6t + 12 (choice A), but remember that the derivative of any constant is zero. The key strategy is to apply the power rule term by term: bring down the exponent as a coefficient and reduce the exponent by one.

This problem tests the basic derivative rules for constants, sums, and differences. To find d'(x) for d(x) = $12x^3$ + $7x^2$ - 4x + 1, apply the power rule to the first term: the derivative of $12x^3$ is $123x^{2}$ = $36x^2$. For the second term, the derivative of $+7x^2$ is $72x^{1}$ = 14x, and for -4x it is $-41x^{0}$ = -4. The constant term +1 has a derivative of 0, so it drops out. A tempting distractor is choice B, which adds +1, but constants vanish in derivatives. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.