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

A function is $S(t)= -14t^2+7t+2$ . What is $S'(t)$ ?
Answer: $-28t+7$ The sum and constant multiple rules under linearity simplify S(t) = - $14t^2$ + 7t + 2. Derivative of - $14t^2$ is -28t, of 7t is 7, of +2 is 0. Thus, S'(t) = -28t + 7. A common misuse is forgetting the negative on quadratic terms. Linear coefficients might be mishandled. For quadratics, use linearity to differentiate powers separately, ignoring constants.
What is the instantaneous rate of change of $f(x) = -x^3 + 5x$ at $x = 2$ ?
Answer: $-7$ The instantaneous rate of change is given by the derivative. First, find the derivative of $f(x)$ : $f'(x) = -3x^2 + 5$ . Then, evaluate the derivative at $x = 2$ : $f'(2) = -3(2)^2 + 5 = -3(4) + 5 = -12 + 5 = -7$ .
A temperature model is $T(t)=5t^6- t^3+4t+7$ . What is $T'(t)$ ?
Answer: $30t^5-3t^2+4$ For the temperature model, linearity rules allow separate handling of terms with sum, difference, and constant multiple. For T(t) = $5t^6$ - $t^3$ + 4t + 7, differentiate each using the power rule. The derivative of $5t^6$ is $30t^5$ , of - $t^3$ is - $3t^2$ , of 4t is 4, and of 7 is 0. Therefore, T'(t) = $30t^5$ - $3t^2$ + 4, which is choice A. Common misuse includes forgetting to multiply by the power for terms like $t^6$ to $t^5$ instead of $6t^5$ . Some derive constants as their value. A transferable strategy is to apply d/dx (c $x^n$ ) = c n $x^{n-1}$ to each term, ignore constants, and use sum rules to assemble the derivative.
At which of the following values of $x$ does the graph of $f(x) = 2x^3 + 3x^2 - 12x$ have a horizontal tangent line?
Answer: $x = 1$ A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative: $f'(x) = 6x^2 + 6x - 12$ . Set the derivative equal to zero and solve for $x$ : $6x^2 + 6x - 12 = 0$ . Divide by 6: $x^2 + x - 2 = 0$ . Factor the quadratic: $(x+2)(x-1) = 0$ . The solutions are $x = -2$ and $x = 1$ . Of the choices given, $x = 1$ is an answer.
Let $f(x) = 3x^2 - 7$ and $g(x) = 5x + 2$ . If $h(x) = 2f(x) - 3g(x) + 4$ , what is $h'(x)$ ?
Answer: $12x - 15$ Using the constant multiple, sum, and difference rules: $h(x) = 2(3x^2 - 7) - 3(5x + 2) + 4 = 6x^2 - 14 - 15x - 6 + 4 = 6x^2 - 15x - 16$ . Therefore, $h'(x) = 12x - 15$ . Choice B incorrectly uses coefficient 6 instead of 12 for the $x$ term. Choice C forgets the $-3g(x)$ term in the constant. Choice D makes both errors.
A distance model is $d(x)= -2x^7-5x^3+9x-1$ . What is $d'(x)$ ?
Answer: $-14x^6-15x^2+9$ For the distance model, linearity rules enable term-by-term application of the power rule with constants and signs. For d(x) = - $2x^7$ - $5x^3$ + 9x - 1, differentiate each term. The derivative of - $2x^7$ is - $14x^6$ , of - $5x^3$ is - $15x^2$ , of 9x is 9, and of -1 is 0. Thus, d'(x) = - $14x^6$ - $15x^2$ + 9, corresponding to choice A. Common misuse includes adding exponents instead of subtracting 1. Some misapply difference by changing signs. A transferable strategy is to handle negatives carefully, use d/dx (c $x^n$ ) = c n $x^{n-1}$ , and sum while dropping constants.
If $y = \sqrt{2}x^2 - \pi x + e + \frac{1}{\sqrt{3}}$ , then $\frac{dy}{dx}$ equals:
Answer: $2\sqrt{2}x - \pi$ Using the power rule and constant rule: $\frac{d}{dx}[\sqrt{2}x^2] = \sqrt{2} \cdot 2x = 2\sqrt{2}x$ , $\frac{d}{dx}[-\pi x] = -\pi$ , and $\frac{d}{dx}[e + \frac{1}{\sqrt{3}}] = 0$ since both $e$ and $\frac{1}{\sqrt{3}}$ are constants. Choice A incorrectly differentiates the constant $\frac{1}{\sqrt{3}}$ . Choice B incorrectly keeps $e$ as non-zero. Choice D uses wrong coefficient for $x^2$ term and keeps constant $e$ .
For which of the following values of $x$ is the tangent line to the graph of $g(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 10x + 1$ parallel to the x-axis?
Answer: $x = 5$ A tangent line parallel to the x-axis has a slope of 0. We need to find where the derivative of $g(x)$ is zero. $g'(x) = x^2 - 3x - 10$ . Set $g'(x) = 0$ : $x^2 - 3x - 10 = 0$ . Factoring the quadratic gives $(x-5)(x+2) = 0$ . The solutions are $x=5$ and $x=-2$ . Of the choices provided, $x=5$ is a correct value.
For $v(t)=6t^8-2t^7+5t^2-9t+1$ , what is $v'(t)$ ?
Answer: $48t^7-14t^6+10t-9$ The linearity rules facilitate term-by-term differentiation for v(t) = $6t^8$ - $2t^7$ + $5t^2$ - 9t + 1. Derivative of $6t^8$ is $48t^7$ , of - $2t^7$ is - $14t^6$ , of $5t^2$ is 10t, of -9t is -9, of +1 is 0. Thus, v'(t) = $48t^7$ - $14t^6$ + 10t - 9. A common misuse is miscalculating coefficients for high powers, like 6*8=48. Signs need precise tracking. For complex polynomials, apply power rule via linearity to each term individually.
What is the slope of the line tangent to the graph of $y = 4x^5 - 7x^3 + 2$ at $x = -1$ ?
Answer: $-1$ The slope of the tangent line is the value of the derivative at the given point. First, find the derivative: $\frac{dy}{dx} = \frac{d}{dx}(4x^5 - 7x^3 + 2) = 20x^4 - 21x^2$ . Now, evaluate the derivative at $x = -1$ : $20(-1)^4 - 21(-1)^2 = 20(1) - 21(1) = 20 - 21 = -1$ .
Let $Y(t)=2t^6-7t^5+9t^2+6$ . What is $Y'(t)$ ?
Answer: $12t^5-35t^4+18t$ Using linearity, compute Y'(t) for Y(t) = $2t^6$ - $7t^5$ + $9t^2$ + 6 term by term. Derivative of $2t^6$ is $12t^5$ , of - $7t^5$ is - $35t^4$ , of $9t^2$ is 18t, of +6 is 0. Thus, Y'(t) = $12t^5$ - $35t^4$ + 18t. A common misuse is not multiplying by exponents for higher powers, like $2t^5$ for $2t^6$ . Constants vanish, but are sometimes retained. For polynomials, leverage sum rules with power rule per term.
Let $q(x)=2x^8+9x^4-3x^2+12$ . What is $q'(x)$ ?
Answer: $16x^7+36x^3-6x$ Linearity rules facilitate term-by-term differentiation for q(x) using sum, difference, and constant multiple. For q(x) = $2x^8$ + $9x^4$ - $3x^2$ + 12, apply the power rule. The derivative of $2x^8$ is $16x^7$ , of $9x^4$ is $36x^3$ , of - $3x^2$ is -6x, and of 12 is 0. Thus, q'(x) = $16x^7$ + $36x^3$ - 6x, corresponding to choice A. A common rule misuse is miscalculating the new coefficient, like 2*8=16 but forgetting for higher powers. Another is keeping the constant in the derivative. A transferable strategy is to calculate each c * n for the coefficient, reduce power by 1, drop constants, and sum with original operations.
A model for elevation is $h(x)=7x^4-3x^2+10x-6$ . What is $h'(x)$ ?
Answer: $28x^3-6x+10$ The derivative of a polynomial is found using the linearity rules, which include the sum rule, difference rule, and constant multiple rule, allowing us to differentiate each term separately. For h(x) = $7x^4$ - $3x^2$ + 10x - 6, we apply the power rule to each term while respecting these rules. The derivative of $7x^4$ is $28x^3$ , of - $3x^2$ is -6x, of 10x is 10, and of -6 is 0. Therefore, h'(x) = $28x^3$ - 6x + 10, corresponding to choice A. A common misuse is forgetting to multiply by the exponent, such as deriving $x^4$ to $x^3$ instead of $4x^3$ . Another error is incorrectly applying the constant rule by not setting the derivative of constants to zero. A transferable rule-selection strategy is to break the polynomial into individual terms, apply the power rule d/dx (c $x^n$ ) = c n $x^{n-1}$ to each, and sum the results while ignoring constants.
Let $y(x)=11x^9+3x^5-x+8$ . What is $y'(x)$ ?
Answer: $99x^8+15x^4-1$ Linearity of differentiation supports separate term handling for y(x) with sum and constant multiple rules. For y(x) = $11x^9$ + $3x^5$ - x + 8, apply the power rule. The derivative of $11x^9$ is $99x^8$ , of $3x^5$ is $15x^4$ , of -x is -1, and of 8 is 0. So, y'(x) = $99x^8$ + $15x^4$ - 1, matching choice A. A common rule misuse is incorrect coefficient calculation for high powers like 11*9=99. Another is deriving -x as 0. A transferable strategy is to compute large coefficients accurately, apply d/dx (c $x^n$ ) = c n $x^{n-1}$ , and assemble using linearity.
A function is $b(x)= -12x^3+3x^2+9x+4$ . What is $b'(x)$ ?
Answer: $-36x^2+6x+9$ Linearity rules allow separate handling of terms in b(x) with differences and multiples. For b(x) = - $12x^3$ + $3x^2$ + 9x + 4, apply the power rule. The derivative of - $12x^3$ is - $36x^2$ , of $3x^2$ is 6x, of 9x is 9, and of 4 is 0. Thus, b'(x) = - $36x^2$ + 6x + 9, matching choice A. A common misuse is misapplying negative multiples. Some derive linear to zero. A transferable strategy is to emphasize negatives, apply d/dx (c $x^n$ ) = c n $x^{n-1}$ , drop constants, and sum.
Let $f(x) = x^4 - 8x^3 + 10x - 1$ . What is $f''(x)$ ?
Answer: $12x^2 - 48x$ To find the second derivative, we first find the first derivative. $f'(x) = 4x^3 - 24x^2 + 10$ . Then, we differentiate $f'(x)$ to find $f''(x)$ . $f''(x) = \frac{d}{dx}(4x^3 - 24x^2 + 10) = 12x^2 - 48x$ .
Let $f$ and $g$ be differentiable functions. If $h(x) = 4f(x) - 3g(x) + 2$ , and it is known that $f'(2) = 5$ and $g'(2) = -1$ , what is the value of $h'(2)$ ?
Answer: $23$ First, find the derivative of $h(x)$ using the sum, difference, and constant multiple rules. $h'(x) = 4f'(x) - 3g'(x) + 0$ . Now, substitute $x=2$ into the expression for $h'(x)$ : $h'(2) = 4f'(2) - 3g'(2)$ . Using the given values, $h'(2) = 4(5) - 3(-1) = 20 + 3 = 23$ .
Let $f(x) = \frac{x^4}{4} - \frac{x^2}{2} + 5$ . For which value of $x$ does $f'(x) = 0$ ?
Answer: $x = 1$ First, find the derivative of $f(x)$ : $f'(x) = \frac{1}{4}(4x^3) - \frac{1}{2}(2x) + 0 = x^3 - x$ . Now, set the derivative equal to zero: $x^3 - x = 0$ . Factor out an $x$ : $x(x^2 - 1) = 0$ . This gives solutions $x=0$ and $x^2=1$ , so $x = -1, 0, 1$ . From the given choices, $x=1$ is a correct answer.
Let $P(x) = ax^3 + bx^2 + cx + d$ where $a$ , $b$ , $c$ , and $d$ are constants. If $\frac{d}{dx}[P(x) + Q(x)] = 6x^2 - 4x + 7$ and $Q'(x) = 2x^2 + x - 3$ , what is $P'(x)$ ?
Answer: $4x^2 - 5x + 10$ When you encounter problems involving derivatives of sums, remember that the derivative of a sum equals the sum of derivatives: $\frac{d}{dx}[P(x) + Q(x)] = P'(x) + Q'(x)$ . This fundamental property is key to solving this problem. Since you know that $\frac{d}{dx}[P(x) + Q(x)] = 6x^2 - 4x + 7$ and $Q'(x) = 2x^2 + x - 3$ , you can substitute into the sum rule: $P'(x) + Q'(x) = 6x^2 - 4x + 7$ . Therefore, $P'(x) + (2x^2 + x - 3) = 6x^2 - 4x + 7$ . Solving for $P'(x)$ : $P'(x) = (6x^2 - 4x + 7) - (2x^2 + x - 3) = 6x^2 - 4x + 7 - 2x^2 - x + 3 = 4x^2 - 5x + 10$ . This confirms answer choice D. Choice A ( $8x^2 - 3x + 10$ ) results from incorrectly adding $P'(x) + Q'(x)$ instead of subtracting $Q'(x)$ . Choice B ( $4x^2 - 5x + 4$ ) shows correct coefficient work for the $x^2$ and $x$ terms but an error in the constant term ( $7 - (-3) = 10$ , not $4$ ). Choice C ( $8x^2 - 3x + 4$ ) combines both mistakes: adding instead of subtracting the polynomial terms AND miscalculating the constant. Always remember: when you have $f'(x) + g'(x) = h'(x)$ and need to find one derivative, isolate it by subtracting the known derivative from both sides. Double-check your arithmetic, especially with signs when subtracting polynomials.
If $y = 7e^x - 4\ln(x)$ , what is $\frac{dy}{dx}$ ?
Answer: $7e^x - \frac{4}{x}$ We use the constant multiple and difference rules. The derivative of $7e^x$ is $7e^x$ . The derivative of $-4\ln(x)$ is $-4(\frac{1}{x}) = -\frac{4}{x}$ . Combining these, we get $\frac{dy}{dx} = 7e^x - \frac{4}{x}$ .

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