Derivative Rules of Constant, Sum, Difference - AP Calculus BC

1. The functions f and g are differentiable. The graph of y = f'(x) is a line of slope 2 passing through (0, 1), and the graph of y = g'(x) is a line of slope -1 passing through (0, 4). Let h(x) = 3f(x) - 2g(x) + 7. What is the value of h''(1)?
   Answer: 8 This question tests your understanding of derivatives of linear combinations and how to find equations of lines from point-slope information. When you're given information about derivatives as lines, start by writing their equations. The graph of $f'(x)$ has slope 2 and passes through $(0,1)$, so $f'(x) = 2x + 1$. Similarly, $g'(x)$ has slope -1 and passes through $(0,4)$, so $g'(x) = -x + 4$. To find $h''(1)$, you need to differentiate $h(x) = 3f(x) - 2g(x) + 7$. Using the linearity of derivatives: $h'(x) = 3f'(x) - 2g'(x)$. Substituting the expressions you found: $h'(x) = 3(2x + 1) - 2(-x + 4) = 6x + 3 + 2x - 8 = 8x - 5$. Taking the derivative again: $h''(x) = 8$. Since the second derivative is constant, $h''(1) = 8$. Choice A (3) represents the coefficient of $f(x)$ in the original function. Choice B (4) is the y-intercept of $g'(x)$, which students might mistakenly think relates to the second derivative. Choice C (7) is the constant term in $h(x)$, but constants disappear when differentiating twice. Remember that when dealing with linear functions, the derivative is the slope, and the second derivative of any linear function is zero. However, when you have combinations of functions, you must carefully apply derivative rules before concluding that second derivatives vanish.
2. A function is defined by $f(x)= -9x^4+7x^3-5$. What is $f'(x)$?
   Answer: $-36x^3+21x^2$ 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.
3. A particle's position is given by $s(t) = -16t^2 + 50t + 5$. What is the instantaneous rate of change of the position with respect to time $t$ at $t=2$?
   Answer: $-14$ The instantaneous rate of change is the derivative of the position function. First, find $s'(t)$: $s'(t) = -16(2t) + 50 = -32t + 50$. Then, evaluate $s'(t)$ at $t=2$: $s'(2) = -32(2) + 50 = -64 + 50 = -14$.
4. The profit model is $P(t)=-4t^8+2t^4-11$; find the marginal profit $P'(t)$.
   Answer: $-32t^7+8t^3$ This problem tests the basic derivative rules for constants, sums, and differences. To find P'(t) for P(t) = -$4t^8$ + $2t^4$ - 11, apply the power rule to the first term: the derivative of -$4t^8$ is -4*8*$t^{7}$ = -$32t^7$. For the second term, the derivative of +$2t^4$ is 2*4*$t^{3}$ = $8t^3$. The constant term -11 has a derivative of 0, so it is excluded. A tempting distractor is choice C, which subtracts -11, but constants do not appear in the derivative. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.
5. A model for a tank's water volume is $V(t)=7t^4-3t^2+12$. What is $V'(t)$?
   Answer: $28t^3-6t$ This problem requires applying basic derivative rules to find the rate of change of water volume. To find V'(t), we differentiate each term separately: the derivative of 7t⁴ is 28t³ (using the power rule: bring down the 4 and reduce the exponent by 1), the derivative of -3t² is -6t, and the derivative of the constant 12 is 0. Combining these results gives V'(t) = 28t³ - 6t + 0 = 28t³ - 6t. A common error would be to include the constant term in the derivative (choice A), but constants always have a derivative of zero. Remember: when differentiating a polynomial, apply the power rule to each term and constants disappear.
6. If $p(q) = \frac{1}{3}q^3 - \frac{1}{2}q^2 + q - \sqrt{2}$, find $p'(q)$.
   Answer: $q^2 - q + 1$ Apply the power rule and sum/difference rules. The derivative of $\frac{1}{3}q^3$ is $\frac{1}{3}(3q^2) = q^2$. The derivative of $-\frac{1}{2}q^2$ is $-\frac{1}{2}(2q) = -q$. The derivative of $q$ is 1. The term $-\sqrt{2}$ is a constant, so its derivative is 0. Combining these gives $p'(q) = q^2 - q + 1$.
7. A company's cost is $C(x)=12x^5-8x^2+6x$ (dollars). What is $C'(x)$?
   Answer: $60x^4-16x+6$ This problem involves finding the marginal cost function by differentiating the cost function. Given C(x) = 12x⁵ - 8x² + 6x, we apply the power rule term by term. The derivative of 12x⁵ is 12·5x⁴ = 60x⁴, the derivative of -8x² is -8·2x = -16x, and the derivative of 6x is 6·1 = 6. Thus, C'(x) = 60x⁴ - 16x + 6. A common mistake would be to write -16x² instead of -16x (choice B), forgetting that the power rule reduces the exponent by one. The systematic approach is to handle each term separately: multiply the coefficient by the exponent, then decrease the exponent by one.
8. A model for pressure is $P(t)=11-6t^5+4t^3$ (kPa). What is $P'(t)$?
   Answer: $-30t^4+12t^2$ This problem requires finding the rate of change of pressure by differentiating. Starting with P(t) = 11 - 6t⁵ + 4t³, we apply the power rule to each term. The derivative of the constant 11 is 0, the derivative of -6t⁵ is -6·5t⁴ = -30t⁴, and the derivative of 4t³ is 4·3t² = 12t². Thus, P'(t) = 0 - 30t⁴ + 12t² = -30t⁴ + 12t². A potential error would be to write 12t³ instead of 12t² (choice D), forgetting that differentiation reduces the exponent by one. Remember that constants differentiate to zero and apply the power rule carefully to each variable term.
9. A temperature model is $T(t)=6t^2-(t^6)+15t$. Find $T'(t)$.
   Answer: $12t-6t^5+15$ 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.
10. A chemical concentration is $c(t)=2t^9-6t^3+t$; what is the rate $c'(t)$?
    Answer: $18t^8-18t^2+1$ This problem tests the basic derivative rules for constants, sums, and differences. To find c'(t) for c(t) = $2t^9$ - $6t^3$ + t, apply the power rule to the first term: the derivative of $2t^9$ is 2*9*$t^{8}$ = $18t^8$. For the second term, the derivative of -$6t^3$ is -6*3*$t^{2}$ = -$18t^2$, and for +t (which is $t^1$) it is 1*$t^{0}$ = 1. There is no constant term here, but the process remains the same for each power. A tempting distractor is choice B, which uses -$6t^2$ instead of -$18t^2$, forgetting to multiply by the exponent 3. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.
11. Let f and h be differentiable functions such that h'(x) = 5 - $\frac{d}{dx}[2f(x)]$. If f(3) = 1 and h(3) = 6, which of the following defines h(x) in terms of f(x)?
    Answer: h(x) = 5x - 2f(x) - 7 This question tests your ability to work backwards from a derivative to find the original function using antiderivatives and initial conditions. You're given that $h'(x) = 5 - \frac{d}{dx}[2f(x)] = 5 - 2f'(x)$. To find $h(x)$, you need to integrate this derivative. When you integrate $h'(x) = 5 - 2f'(x)$, you get $h(x) = 5x - 2f(x) + C$, where $C$ is the constant of integration. To find $C$, use the given initial conditions: $f(3) = 1$ and $h(3) = 6$. Substituting into your general form: $6 = 5(3) - 2(1) + C$, which gives you $6 = 15 - 2 + C = 13 + C$. Therefore, $C = -7$. This means $h(x) = 5x - 2f(x) - 7$, which is choice A. Choice B ($h(x) = 5x - 2f(x)$) forgets to include the constant of integration, giving $C = 0$ instead of $C = -7$. Choice C ($h(x) = -2f(x) + 8$) incorrectly integrates the constant term 5, treating it as if it contributes nothing to the antiderivative rather than $5x$. Choice D ($h(x) = 5x - 2f(x) + 19$) makes an arithmetic error when solving for the constant, possibly calculating $C = 6 + 15 + 2 = 23$ instead of $C = 6 - 15 + 2 = -7$. Study tip: When finding a function from its derivative, always integrate completely (including the constant) and use initial conditions to solve for that constant. Double-check your arithmetic in the final substitution step.
12. The position of a cart is $s(x)=-5x^3+4x-9$ meters; find the velocity $s'(x)$.
    Answer: $-15x^2+4$ 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 $-5*3*x^{2} = -15x^2$. For the second term, the derivative of $+4x$ is $4*1*x^{0} = 4$. 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.
13. The cost function is $C(x)=15x^2-6x-11$. What is $C'(x)$?
    Answer: $30x-6$ This marginal cost problem requires applying basic derivative rules to find C'(x). Starting with C(x) = 15x² - 6x - 11, we differentiate each term: the derivative of 15x² is 30x, the derivative of -6x is -6, and the derivative of the constant -11 is 0. Thus C'(x) = 30x - 6 + 0 = 30x - 6. Choice D incorrectly includes the constant -11 in the derivative, but constants always disappear when differentiating. The strategy is to handle each term separately using the appropriate rule.
14. For $p(x)=11x^3-5x^3+9x-6$, what is the derivative $p'(x)$?
    Answer: $18x^2+9$ This problem requires simplifying before differentiating to avoid errors. Given p(x) = 11x³ - 5x³ + 9x - 6, we should first combine like terms: p(x) = 6x³ + 9x - 6. Now differentiating: 6x³ becomes 6·3x² = 18x²; 9x becomes 9·1 = 9; and -6 becomes 0. Thus, p'(x) = 18x² + 9 + 0 = 18x² + 9. Choice B shows 6x² + 9, which would result from forgetting to multiply by the exponent 3 when differentiating 6x³. Always simplify algebraically first when possible, then apply derivative rules to the simplified expression.
15. A tank's water volume is modeled by $V(t)=7t^4-3t^2+12$. What is $V'(t)$?
    Answer: $28t^3-6t$ 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.
16. Let $f(x) = \frac{x^4}{4} - 2e^x + \ln(5)$. What is $f'(x)$?
    Answer: $x^3 - 2e^x$ Differentiate term by term. The derivative of $\frac{x^4}{4}$ is $\frac{1}{4}(4x^3) = x^3$. The derivative of $-2e^x$ is $-2e^x$. The term $\ln(5)$ is a constant, so its derivative is 0. Therefore, $f'(x) = x^3 - 2e^x$.
17. Let f and g be differentiable functions. What is $\lim_{h \to 0} \frac{(5f(x+h) - 2g(x+h)) - (5f(x) - 2g(x))}{h}$?
    Answer: 5f'(x) - 2g'(x) When you see a limit expression of the form $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h}$, you're looking at the definition of a derivative. This is the fundamental building block of differential calculus, so recognizing this pattern is crucial. Let's work through this systematically. We have: $\lim_{h \to 0} \frac{(5f(x+h) - 2g(x+h)) - (5f(x) - 2g(x))}{h}$ Rearranging the numerator by grouping like terms: $\lim_{h \to 0} \frac{5f(x+h) - 5f(x) - 2g(x+h) + 2g(x)}{h}$ Factor out the constants: $\lim_{h \to 0} \frac{5(f(x+h) - f(x)) - 2(g(x+h) - g(x))}{h}$ Using limit properties, we can split this: $5 \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} - 2 \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$ Each limit is now in the derivative definition form, giving us $5f'(x) - 2g'(x)$. Choice A ($5f(x) - 2g(x)$) forgets to take the derivative entirely. Choice C ($3(f'(x) - g'(x))$) incorrectly combines the coefficients and loses track of the individual derivatives. Choice D is wrong because both f and g are given to be differentiable, guaranteeing these limits exist. Study tip: Whenever you see $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h}$, immediately think "derivative of F." The linearity of derivatives means constants factor out and addition/subtraction distribute: $(af + bg)' = af' + bg'$.
18. A tank's water volume is modeled by $V(t)=12-5t+3t^2$. What is $V'(t)$?
    Answer: $-5+6t$ This problem requires applying basic derivative rules to find the rate of change of water volume. To find V'(t) from V(t) = 12 - 5t + 3t², we differentiate each term separately: the derivative of the constant 12 is 0, the derivative of -5t is -5, and the derivative of 3t² is 6t. Combining these results gives V'(t) = 0 - 5 + 6t = -5 + 6t. A common error would be to include the constant 12 in the derivative (as in choice C), but constants always have derivative zero. The key strategy is to differentiate each term independently and then combine the results.
19. Let $f$ be a differentiable function and let $c$ be a real number constant. If $g(x) = f(x) + f(c)$, then $g'(x)=$
    Answer: $f'(x)$ We use the sum rule for derivatives: $g'(x) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(f(c))$. The derivative of $f(x)$ is $f'(x)$. Since $c$ is a constant, $f(c)$ is also a constant value. The derivative of any constant is 0. Therefore, $g'(x) = f'(x) + 0 = f'(x)$.
20. Find $D_x(3x^3 - 2\cos(x) + e^2)$.
    Answer: $9x^2 + 2\sin(x)$ Differentiate term by term. The derivative of $3x^3$ is $9x^2$. The derivative of $-2\cos(x)$ is $-2(-\sin(x)) = 2\sin(x)$. The term $e^2$ is a constant, so its derivative is 0. Thus, the result is $9x^2 + 2\sin(x)$.

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