Derivative Rules of Constant, Sum, Difference
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AP Calculus AB › Derivative Rules of Constant, Sum, Difference
Let $f(t)=9t^3-2t^2-6t+1$. What is $f'(t)$?
$27t^2-4t-6$
$27t^3-4t^2-6t$
$9t^3-2t^2-6t+1$
$27t^2-4t-5$
$9t^2-2t-6$
Explanation
Linearity rules allow us to handle the sum and differences in f(t) by differentiating individually with the constant multiple rule. For f(t) = $9t^3$ - $2t^2$ - 6t + 1, apply the power rule to every term. The derivative of $9t^3$ is $27t^2$, of $-2t^2$ is -4t, of -6t is -6, and of 1 is 0. Therefore, f'(t) = $27t^2$ - 4t - 6, corresponding to choice A. A common rule misuse is applying the power rule without multiplying by the exponent, e.g., $t^3$ to $t^2$ instead of $3t^2$. Another is ignoring negative signs in differences. A transferable strategy is to process each term with d/dx (c $x^n$) = c n $x^{n-1}$, remember constants vanish, and maintain sign integrity in the final expression.
Let $F(x)=3x^3-4x^2+5x-6$. What is the derivative $F'(x)$?
$9x^3-8x^2+5x$
$3x^3-4x^2+5x-6$
$9x^2-8x-6$
$3x^2-8x+5$
$9x^2-8x+5$
Explanation
Linearity rules allow independent differentiation of terms in F(x) = $3x^3$ - $4x^2$ + 5x - 6. Derivative of $3x^3$ is $9x^2$, of $-4x^2$ is -8x, of 5x is 5, of -6 is 0. Thus, F'(x) = $9x^2$ - 8x + 5. A common misuse is subtracting instead of adding derivatives in sums. Constants are often incorrectly included. Decompose using linearity and apply power rule to each term for accuracy.
Let $G(x)= -6x^2+13x-15$. What is the derivative $G'(x)$?
$-6x+13$
$-6x^2+13x-15$
$-12x-15$
$-12x^2+13x$
$-12x+13$
Explanation
The sum and difference rules of linearity allow term-by-term differentiation of polynomials. For G(x) = $-6x^2$ + 13x - 15, derivative of $-6x^2$ is -12x, of 13x is 13, and of -15 is 0. Thus, G'(x) = -12x + 13. A common misuse is including the constant's derivative as nonzero, like adding -15. Signs on linear terms must be preserved accurately. For quadratic functions, use linearity to differentiate each power separately, ensuring constants are ignored.
A function is $y(t)=13t^2+4t+9$. What is $y'(t)$?
$26t+4$
$13t^2+4t+9$
$26t^2+4t$
$26t+9$
$13t+4$
Explanation
Linearity rules facilitate term-by-term differentiation for y(t) = $13t^2$ + 4t + 9. Derivative of $13t^2$ is 26t, of 4t is 4, of +9 is 0. Thus, y'(t) = 26t + 4. A common misuse is including the constant in the derivative, like adding 9. Linear terms are sometimes overlooked. For quadratics, apply sum rules and power rule individually to each term.
Let $N(x)=5x^4+7x^3-14x+9$. What is $N'(x)$?
$20x^4+21x^3-14x$
$5x^4+7x^3-14x+9$
$20x^3+21x^2+9$
$5x^3+21x^2-14$
$20x^3+21x^2-14$
Explanation
Linearity rules facilitate term-by-term differentiation for N(x) with sums and multiples. For N(x) = $5x^4$ + $7x^3$ - 14x + 9, use the power rule. The derivative of $5x^4$ is $20x^3$, of $7x^3$ is $21x^2$, of -14x is -14, and of 9 is 0. Therefore, N'(x) = $20x^3$ + $21x^2$ - 14, which is choice A. A common misuse is forgetting multiplication for cubic terms. Some include constants. A transferable strategy is to sequence terms, apply d/dx (c $x^n$) = c n $x^{n-1}$, and combine using linearity.
A rate function is $r(t)= -9t^4+10t^2+7t-3$. What is $r'(t)$?
$-36t^3+20t-3$
$-36t^3+20t+7$
$-9t^4+10t^2+7t-3$
$-36t^4+20t^2+7t$
$-9t^3+20t+7$
Explanation
The rate function uses linearity rules to differentiate terms independently with differences and multiples. For r(t) = $-9t^4$ + $10t^2$ + 7t - 3, use the power rule. The derivative of $-9t^4$ is $-36t^3$, of $10t^2$ is 20t, of 7t is 7, and of -3 is 0. Therefore, r'(t) = $-36t^3$ + 20t + 7, which is choice A. Common misuse is forgetting to multiply by the power for even degrees. Some keep the constant. A transferable strategy is to apply d/dx (c $x^n$) = c n $x^{n-1}$ to each, preserve signs from differences, and ignore constants in the sum.
For $S(x)=10x^5-5x^4+4x^3-1$, what is $S'(x)$?
$50x^4-5x^3+12x^2$
$50x^4-20x^3+12x^2$
$10x^5-5x^4+4x^3-1$
$50x^5-20x^4+12x^3$
$10x^4-20x^3+12x^2$
Explanation
The linearity of differentiation allows us to compute the derivative of a polynomial by differentiating each term separately and respecting constant multiples. For S(x) = $10x^5$ - $5x^4$ + $4x^3$ - 1, apply the power rule to each term: the derivative of $ax^n$ is a n $x^{n-1}$. Thus, the derivative of $10x^5$ is $50x^4$, of $-5x^4$ is $-20x^3$, of $4x^3$ is $12x^2$, and the constant -1 has derivative 0. Therefore, S'(x) = $50x^4$ - $20x^3$ + $12x^2$. A common misuse is forgetting that the derivative of a constant is zero, leading to incorrectly including terms like -1 in the derivative. Another error is mishandling the constant multiple, such as not multiplying the coefficient by the exponent. For differentiating polynomials, always break down into individual terms, apply the power rule with linearity, and confirm constants vanish.
Let $y(x)=11x^9+3x^5-x+8$. What is $y'(x)$?
$11x^8+15x^4-1$
$99x^9+15x^5-x$
$99x^8+15x^4-1$
$11x^9+3x^5-x+8$
$99x^8+15x^5-1$
Explanation
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.
Let $f(x)= -2x^4-9x^3+6x^2-1$. What is $f'(x)$?
$-2x^4-9x^3+6x^2-1$
$-8x^4-27x^3+12x^2$
$-2x^3-27x^2+12x$
$-8x^3-27x^2-1$
$-8x^3-27x^2+12x$
Explanation
Using linearity, compute f'(x) for f(x) = $-2x^4$ - $9x^3$ + $6x^2$ - 1 by differentiating each term. Derivative of $-2x^4$ is $-8x^3$, of $-9x^3$ is $-27x^2$, of $6x^2$ is 12x, of -1 is 0. Thus, f'(x) = $-8x^3$ - $27x^2$ + 12x. A common misuse is ignoring negative signs, turning terms positive incorrectly. Constants are often mistakenly differentiated. Break polynomials into sums and use power rule per term with linearity for reliable outcomes.
For $m(t)=6t^2-15t+9$, what is the derivative $m'(t)$?
$12t^2-15t$
$6t^2-15t+9$
$12t-15$
$12t-6$
$6t-15$
Explanation
Linearity rules allow independent differentiation of terms in m(t) using sum, difference, and constant multiple. For m(t) = $6t^2$ - 15t + 9, apply the power rule. The derivative of $6t^2$ is 12t, of -15t is -15, and of 9 is 0. Therefore, m'(t) = 12t - 15, which is choice A. A common misuse is treating quadratic terms as linear without multiplying by 2. Another error is deriving the constant as non-zero. A transferable strategy is to focus on low-degree polynomials by applying d/dx (c $x^n$) = c n $x^{n-1}$ term-wise and combining via linearity.