Applying the Power Rule - AP Calculus BC

1. The cost function is $C(x)=12x^4-5x^3+2x+11$. What is $C'(x)$?
   Answer: $48x^3-15x^2+2$ To find C'(x), we apply the power rule to each term of the cost function C(x) = $12x^4$ - $5x^3$ + 2x + 11. Using the power rule d/dx($x^n$) = $nx^{n-1}$, we get: the derivative of $12x^4$ is 4·$12x^3$ = $48x^3$, the derivative of -$5x^3$ is 3·(-5)$x^2$ = -$15x^2$, the derivative of 2x is 2·$1x^0$ = 2, and the derivative of the constant 11 is 0. Choice C ($12x^3$ - $5x^2$ + 2) makes the common error of only bringing down the exponent without multiplying by the original coefficient. The key to the power rule is to multiply the coefficient by the exponent before reducing the exponent by 1.
2. A particle's velocity is modeled by $v(t)=12t^{1/2}-3t^{3/2}$. What is $v'(t)$?
   Answer: $6t^{-1/2}-\frac{9}{2}t^{1/2}$ The power rule is the key skill here, which states that the derivative of a term like $a t^n$ is $a n t^{n-1}$, including for fractional exponents. To find $v'(t)$ from $v(t) = 12t^{1/2} - 3t^{3/2}$, begin with the first term: the derivative of $12t^{1/2}$ is $12*(1/2) t^{-1/2} = 6t^{-1/2}$. Then, the derivative of $-3t^{3/2}$ is $-3*(3/2) t^{1/2} = - (9/2) t^{1/2}$. This results in $6t^{-1/2} - (9/2) t^{1/2}$. A tempting distractor like choice E, $6t^{-1/2} - (3/2) t^{1/2}$, fails by not multiplying the -3 by the full exponent 3/2. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
3. The revenue function is $R(x)=5x^6-2x^3+11$. What is $R'(x)$?
   Answer: $30x^5-6x^2$ To find R'(x), we need to apply the power rule to each term of the revenue function. The power rule tells us that the derivative of $x^n$ is $nx^{n-1}$. For R(x) = $5x^6$ - $2x^3$ + 11, we differentiate term by term: $5x^6$ becomes $30x^5$, -$2x^3$ becomes -$6x^2$, and the constant 11 becomes 0. Choice A incorrectly includes the constant 11 in the derivative, which is wrong because the derivative of any constant is zero. When applying the power rule, always remember to drop constant terms from your final answer.
4. The position of a cart is $s(t)=-4t^3+2t^2-10$; which expression equals the velocity $s'(t)$?
   Answer: $-12t^2+4t$ This problem applies the power rule to find the velocity from the position function s(t) = -$4t^3$ + $2t^2$ - 10. Differentiate -$4t^3$ to get -4*3 $t^2$ = -$12t^2$. For $2t^2$, it's 2*2 $t^1$ = 4t. The constant -10 has a derivative of 0, with no other terms. Choice B distracts by not reducing the exponents properly, leading to terms like $t^3$. A general power rule strategy is to process each polynomial term by multiplying its coefficient by the exponent and subtracting one from the exponent, ensuring constants vanish.
5. For $g(x)=12x^3-5x+7x^2$, what is $g'(x)$?
   Answer: $36x^2-5+14x$ Finding g'(x) requires applying the power rule to each term, but first we should rewrite g(x) = $12x^3$ - 5x + $7x^2$ in standard form. The power rule states that the derivative of $x^n$ is $nx^{n-1}$. Differentiating term by term: $12x^3$ becomes $36x^2$, $7x^2$ becomes 14x, and -5x becomes -5. Choice B incorrectly calculates the derivative of $7x^2$ as 7x instead of 14x, forgetting to multiply by the exponent. When applying the power rule, always multiply the coefficient by the current exponent before reducing the exponent by 1.
6. The function $f(x)=12x^{1/2}+6x^3-4$ models a quantity; what is $f'(x)$?
   Answer: $6x^{-1/2}+18x^2$ This problem uses the power rule for $f(x) = 12x^{1/2} + 6x^3 - 4$, which includes a fractional exponent. Apply it to $12x^{1/2}$, getting $12*(1/2) x^{-1/2} = 6x^{-1/2}$. For $6x^3$, it's $6*3 x^2 = 18x^2$. The constant -4 differentiates to 0. Choice C fails by not subtracting 1 from the 1/2 exponent, leaving a positive power. Always apply the power rule consistently by multiplying by the exponent and reducing it by one for every term, regardless of whether exponents are integers or fractions.
7. A polynomial model for demand is $D(x)=x^8-4x^5+2x^2$; find $D'(x)$.
   Answer: $8x^7-20x^4+4x$ This problem applies the power rule to the demand model D(x) = $x^8$ - $4x^5$ + $2x^2$. Differentiate $x^8$ to 1*8 $x^7$ = $8x^7$. For -$4x^5$, it's -4*5 $x^4$ = -$20x^4$. Then, $2x^2$ becomes 2*2 $x^1$ = 4x, with no constant. Choice B errs by not reducing exponents, retaining $x^8$. A transferable approach is to treat each term separately in the power rule, multiplying by the exponent and subtracting one, ensuring no constants affect the derivative.
8. A model for pressure is $P(x)=-7x^3+2x^2-9x+1$. What is $P'(x)$?
   Answer: $-21x^2+4x-9$ The power rule is the key skill here, which states that the derivative of a term like a $x^n$ is a n $x^{n-1}$, and constants have a derivative of zero. To find P'(x) from P(x) = -$7x^3$ + $2x^2$ - 9x + 1, start with the first term: the derivative of -$7x^3$ is -7*3 $x^2$ = -$21x^2$. Next, the derivative of $2x^2$ is 2*2 $x^1$ = 4x, and the derivative of -9x is -9*1 $x^0$ = -9. The constant +1 differentiates to 0, so the result is -$21x^2$ + 4x - 9. A tempting distractor like choice D, -$21x^2$ + 2x - 9, fails because it forgets to multiply the 2 by the exponent 2 for the second term. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
9. A tank's volume is modeled by $V(t)=t^4-6t^2+t+12$. What is $V'(t)$?
   Answer: $4t^3-12t+1$ Finding V'(t) requires applying the power rule to differentiate the volume function. The power rule states that $\frac{d}{dx} [x^n] = n x^{n-1}$. For $ V(t) = $t^4$ - $6t^2$ + t + 12 $, we differentiate term by term: the derivative of $ $t^4$ $ is $ $4t^3$ $, the derivative of $ -$6t^2$ $ is $ -12t $, the derivative of $ t $ is $ 1 $, and the derivative of $ 12 $ is $ 0 $. Choice C incorrectly omits the coefficient 4 from the first term and halves the coefficient of the second term, showing incomplete application of the power rule. Always multiply the original coefficient by the exponent when applying the power rule.
10. A cost function is $C(x)=9x^4-2x^3+x-12$. What is $C'(x)$?
    Answer: $36x^3-6x^2+1$ The power rule is the key skill here, which states that the derivative of a term like a $x^n$ is a n $x^{n-1}$, and constants have a derivative of zero. To find C'(x) from C(x) = $9x^4$ - $2x^3$ + x - 12, begin with the first term: the derivative of $9x^4$ is 9*4 $x^3$ = $36x^3$. Then, the derivative of -$2x^3$ is -2*3 $x^2$ = -$6x^2$, and the derivative of x (which is $1x^1$) is 1*1 $x^0$ = 1. The constant -12 differentiates to 0, resulting in $36x^3$ - $6x^2$ + 1. A tempting distractor like choice D, $36x^3$ - $2x^2$ + 1, fails by not multiplying the -2 coefficient by its exponent of 3. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
11. A model for water volume is $V(t)=t^6-8t^3+6$. What is $V'(t)$?
    Answer: $6t^5-24t^2$ Finding V'(t) requires applying the power rule to the volume function V(t) = $t^6$ - $8t^3$ + 6. The power rule tells us that the derivative of $t^n$ is $nt^{n-1}$. Applying this to each term: the derivative of $t^6$ is $6t^5$, the derivative of -$8t^3$ is 3·(-8)$t^2$ = -$24t^2$, and the derivative of the constant 6 is 0. Choice D ($6t^5$ - $8t^2$) incorrectly differentiates -$8t^3$ by only reducing the exponent without multiplying by 3, getting -$8t^2$ instead of -$24t^2$. When using the power rule, always remember to multiply the existing coefficient by the exponent before decreasing the exponent by 1.
12. The position of a cart is modeled by $s(t)=4t^5-3t^2+7t-9$. What is $s'(t)$?
    Answer: $20t^4-6t+7$ The power rule is the key skill here, which states that the derivative of a term like a $t^n$ is a n $t^{n-1}$, and constants have a derivative of zero. To find s'(t) from s(t) = $4t^5$ - $3t^2$ + 7t - 9, start with the first term: the derivative of $4t^5$ is 4*5 $t^4$ = $20t^4$. Next, the derivative of -$3t^2$ is -3*2 $t^1$ = -6t, and the derivative of 7t is 7*1 $t^0$ = 7. The constant -9 differentiates to 0, so putting it together gives $20t^4$ - 6t + 7. A tempting distractor like choice D, $20t^4$ - $3t^2$ + 7, fails because it forgets to multiply the coefficients by the exponents and reduce the powers. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
13. The displacement is $d(t)=7t^{-2}-3t^4+2$. What is $d'(t)$?
    Answer: $-14t^{-3}-12t^3$ To find d'(t), we apply the power rule to each term, including the negative exponent. The power rule states that d/dt[$t^n$] = $nt^{n-1}$ for any real n. For d(t) = $7t^{-2}$ - $3t^4$ + 2, we get: the derivative of $7t^{-2}$ is (-2)·$7t^{-3}$ = -$14t^{-3}$, the derivative of -$3t^4$ is 4·(-3)$t^3$ = -$12t^3$, and the derivative of 2 is 0. Choice C incorrectly makes the first term positive, likely forgetting that (-2)·7 = -14, not 14. When differentiating negative exponents, carefully track the signs: the exponent -2 contributes a negative when multiplied by the positive coefficient.
14. A cost function is $C(x)=9x^{1/2}+4x^3-10$. What is $C'(x)$?
    Answer: $\dfrac{9}{2}x^{-1/2}+12x^2$ Finding C'(x) requires applying the power rule, including to fractional exponents. The power rule works for any real exponent: $\frac{d}{dx} [x^n] = n x^{n-1}$. For $ C(x) = $9x^{1/2}$ + $4x^3$ - 10 $, we get: the derivative of $ $9x^{1/2}$ $ is $\frac{1}{2} \cdot 9 x^{-1/2} = \frac{9}{2} x^{-1/2}$, the derivative of $ $4x^3$ $ is $3 \cdot 4 x^2 = 12 x^2$, and the derivative of -10 is 0. Choice B incorrectly adds 1 to the exponent 1/2 instead of subtracting 1, getting $ $x^{1/2}$ $ instead of $ $x^{-1/2}$ $. When applying the power rule to fractional exponents, subtract 1 from the exponent just as with whole numbers.
15. A polynomial model for revenue is $R(x)=x^7-4x^5+6x^2$. What is $R'(x)$?
    Answer: $7x^6-20x^4+12x$ The power rule is the key skill here, which states that the derivative of a term like a $x^n$ is a n $x^{n-1}$. To find R'(x) from R(x) = $x^7$ - $4x^5$ + $6x^2$, begin with the first term: the derivative of $x^7$ (which is $1x^7$) is 1*7 $x^6$ = $7x^6$. Then, the derivative of -$4x^5$ is -4*5 $x^4$ = -$20x^4$, and the derivative of $6x^2$ is 6*2 $x^1$ = 12x. Combining these gives $7x^6$ - $20x^4$ + 12x. A tempting distractor like choice D, $7x^6$ - $4x^4$ + 12x, fails by forgetting to multiply the -4 by the exponent 5. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
16. A tank's volume is $V(t)=t^6-6t^4+9t^2+12$; what is the instantaneous rate $V'(t)$?
    Answer: $6t^5-24t^3+18t$ This problem uses the power rule to compute the instantaneous rate of change for V(t) = $t^6$ - $6t^4$ + $9t^2$ + 12. Apply the rule to $t^6$, yielding 1*6 $t^5$ = $6t^5$. For -$6t^4$, it's -6*4 $t^3$ = -$24t^3$. Then, $9t^2$ becomes 9*2 $t^1$ = 18t, and the constant 12 differentiates to 0. Choice B is tempting but incorrect as it fails to reduce the exponents, keeping terms like $t^6$. Always remember to use the power rule by multiplying the coefficient by the exponent and decreasing the exponent by one for each term, treating constants as zero in the derivative.
17. A model for energy is $E(x)=7x^4-x^3+5x-2$; find $E'(x)$.
    Answer: $28x^3-3x^2+5$ This problem involves the power rule to differentiate the energy model E(x) = $7x^4$ - $x^3$ + 5x - 2. Begin with $7x^4$, deriving 7*4 $x^3$ = $28x^3$. Next, -$x^3$ becomes -1*3 $x^2$ = -$3x^2$. Then, 5x yields 5*1 $x^0$ = 5, and -2 differentiates to 0. Choice C is incorrect because it doesn't decrease the exponents, resulting in $x^4$ terms. For effective use of the power rule, independently multiply each term's coefficient by its exponent and reduce the exponent by one, remembering constants disappear.
18. The energy is modeled by $E(t)=\frac{5}{2}t^8-6t^4+10$. What is $E'(t)$?
    Answer: $20t^7-24t^3$ The power rule is the key skill here, which states that the derivative of a term like a $t^n$ is a n $t^{n-1}$, and constants have a derivative of zero. To find E'(t) from E(t) = (5/2) $t^8$ - 6 $t^4$ + 10, begin with the first term: the derivative of (5/2) $t^8$ is (5/2)*8 $t^7$ = 20 $t^7$. Then, the derivative of -6 $t^4$ is -6*4 $t^3$ = -24 $t^3$. The constant +10 differentiates to 0, giving 20 $t^7$ - 24 $t^3$. A tempting distractor like choice D, 20 $t^7$ - 6 $t^3$, fails by not multiplying the -6 by the exponent 4. Remember, a transferable strategy for the power rule is to handle each term independently, multiply by the exponent, and subtract one from the exponent, ensuring constants vanish.
19. A particle's position is $s(t)=4t^5-3t^2+7t-9$ meters. What is $s'(t)$?
    Answer: $20t^4-6t+7$ This problem requires applying the power rule to find the derivative of the position function. The power rule states that if f(x) = $x^n$, then f'(x) = $nx^{n-1}$. For s(t) = $4t^5$ - $3t^2$ + 7t - 9, we differentiate each term: the derivative of $4t^5$ is 5·$4t^4$ = $20t^4$, the derivative of -$3t^2$ is 2·(-3)$t^1$ = -6t, the derivative of 7t is 7·$1t^0$ = 7, and the derivative of the constant -9 is 0. Choice C incorrectly keeps the original exponents instead of reducing them by 1, showing a misunderstanding of the power rule. Remember: multiply by the exponent, then reduce the exponent by 1 for each term.
20. The function $q(x)=7x^4-9x^2+2x-6$ models profit. What is $q'(x)$?
    Answer: $28x^3-18x+2$ Finding q'(x) requires applying the power rule to model the rate of change in profit. The power rule states that d/dx[$x^n$] = $nx^{n-1}$. For q(x) = $7x^4$ - $9x^2$ + 2x - 6, we differentiate term by term: $7x^4$ becomes $28x^3$, -$9x^2$ becomes -18x, 2x becomes 2, and the constant -6 becomes 0. Choice D incorrectly calculates the derivative of -$9x^2$ as -9x instead of -18x, forgetting to multiply by the exponent 2. To master the power rule, always follow the pattern: multiply by the exponent, then reduce the exponent by 1.

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