Chain Rule - AP Calculus BC

A cost model is $C(x)=\ln\big(7-2x^3\big)$ ; what is $C'(x)$ ?
Answer: $\dfrac{-6x^2}{7-2x^3}$ This problem requires the chain rule to differentiate a composite function. The outer function is the natural logarithm, while the inner function is $7 - 2x^3$ . To differentiate, first take the derivative of the outer function, which is one over the inner function. Then, multiply by the derivative of the inner function, which is $-6x^2$ , resulting in $\frac{-6x^2}{7 - 2x^3}$ . A tempting distractor like choice D has the wrong sign by forgetting the negative from the inner derivative. Always look for compositions where a function is plugged into another, such as a polynomial inside a logarithm, to recognize when to apply the chain rule.
If $g(x)=\tan\!\big(\pi x^2\big)$ models an angle, what is $g'(x)$ ?
Answer: $2\pi x\sec^2(\pi x^2)$ This problem requires the chain rule to differentiate a composite trigonometric function. The function g(x) = tan(πx²) has an outer function f(u) = tan(u) and an inner function u = πx². By the chain rule, g'(x) = sec²(u) · u'(x) = sec²(πx²) · 2πx = 2πx sec²(πx²). Option B shows 2x sec²(πx²), forgetting to include the π factor from differentiating πx². When the inner function contains a constant multiplier, that constant must be included in the final derivative through the chain rule multiplication.
A signal is $I(t)=\sin\big(4t^3-t\big)$ . What is $I'(t)$ ?
Answer: $(12t^2-1)\cos(4t^3-t)$ This problem requires the chain rule to differentiate I(t) = sin(4t³ - t). The outer function is sin(u) and the inner function is u = 4t³ - t. By the chain rule, we differentiate the outer function to get cos(u), then multiply by the derivative of the inner function, which is 12t² - 1. Therefore, I'(t) = cos(4t³ - t) · (12t² - 1) = (12t² - 1)cos(4t³ - t). Choice A incorrectly provides only cos(4t³ - t), forgetting to multiply by the derivative of the inner function. When differentiating trigonometric functions with composite arguments, always remember to multiply by the derivative of the argument.
A response is $R(x)=\left(1+\cos x\right)^7$ . What is $R'(x)$ ?
Answer: $-7\sin x\left(1+\cos x\right)^6$ This problem requires the chain rule to differentiate R(x) = (1 + cos x)⁷. The chain rule states that for composite functions, we multiply the derivative of the outer function by the derivative of the inner function. The outer function is u⁷ with derivative 7u⁶, and the inner function is u = 1 + cos x with derivative -sin x. Applying the chain rule: R'(x) = 7(1 + cos x)⁶ · (-sin x) = -7sin x(1 + cos x)⁶. Choice C incorrectly has 7sin x(1 + cos x)⁶, missing the negative sign from the derivative of cos x. Remember that the derivative of cos x is -sin x, and this negative sign must be included when applying the chain rule.
A utility function is $U(x)=\left(\ln(3x-2)\right)^2$ . What is $U'(x)$ ?
Answer: $\dfrac{6\ln(3x-2)}{3x-2}$ This problem requires the chain rule to differentiate U(x) = (ln(3x - 2))². The chain rule tells us to work from outside to inside, multiplying derivatives at each layer. The outermost function is u² with derivative 2u, the middle function is ln(v) with derivative 1/v, and the innermost function is v = 3x - 2 with derivative 3. Applying the chain rule: U'(x) = 2·ln(3x - 2) · (1/(3x - 2)) · 3 = 6ln(3x - 2)/(3x - 2). Choice A incorrectly gives 2ln(3x - 2)/(3x - 2), forgetting to multiply by 3 (the derivative of 3x - 2). When dealing with nested functions, trace through each layer carefully, multiplying all the derivatives together.
A signal's intensity is $I(x)=\tan\big(5x^3-x\big)$ ; what is $I'(x)$ ?
Answer: $\big(15x^2-1\big)\big(1+\tan^2(5x^3-x)\big)$ To find $I'(x)$ for $I(x)=\tan(5x^3-x)$ , we need the chain rule. The outer function is $\tan(u)$ with derivative $\sec^2(u)=1+\tan^2(u)$ , and the inner function is $u=5x^3-x$ with derivative $15x^2-1$ . Applying the chain rule gives $I'(x)=(15x^2-1)(1+\tan^2(5x^3-x))$ . Choice A incorrectly uses $\tan$ instead of $\sec^2$ as the derivative of tangent. Remember that $\frac{d}{dx}[\tan(u)]=\sec^2(u)\cdot u'=(1+\tan^2(u))\cdot u'$ when differentiating composite trigonometric functions.
A model for pressure is $P(t)=\big(\cos(3t)\big)^5$ ; what is $P'(t)$ ?
Answer: $-15\big(\cos(3t)\big)^4\sin(3t)$ This problem requires the chain rule to differentiate a composite function. The outer function is raising to the fifth power, while the inner function is cos(3t). To differentiate, first take the derivative of the outer function, which is 5 times the inner raised to the fourth power. Then, multiply by the derivative of the inner function, which is -sin(3t) times 3 or -3 sin(3t), resulting in -15 (cos(3t))⁴ sin(3t). A tempting distractor like choice A forgets the negative sign and the factor of 3 from the inner derivative. Always look for compositions where a function is plugged into another, such as a trigonometric function inside a power, to recognize when to apply the chain rule.
A car's speed is $v(t)=\sqrt{2t^3+1}$ . What is $\dfrac{dv}{dt}$ ?
Answer: $\dfrac{6t^2}{2\sqrt{2t^3+1}}$ This problem requires the chain rule to differentiate a square root function. The function v(t) = √(2t³ + 1) can be rewritten as v(t) = $(2t³ + 1)^{1/2}$ , with outer function f(u) = $u^{1/2}$ and inner function u = 2t³ + 1. By the chain rule, v'(t) = f'(u) · u'(t) = (1/2) $u^{-1/2}$ · 6t² = (1/2) $(2t³ + 1)^{-1/2}$ · 6t² = 6t²/[2√(2t³ + 1)]. Answer B incorrectly omits the factor of 1/2 from the power rule, giving 3t²/√(2t³ + 1) instead. When you see a radical or fractional exponent with a non-trivial expression inside, the chain rule is needed.
A growth model is $G(x)=e^{2x^2-3x}$ . What is $G'(x)$ ?
Answer: $(4x-3)e^{2x^2-3x}$ This problem requires the chain rule to differentiate G(x) = $e^{2x² - 3x}$ . The outer function is $e^u$ and the inner function is u = 2x² - 3x. Applying the chain rule, we differentiate the outer function to get $e^u$ (since the derivative of $e^u$ is itself), then multiply by the derivative of the inner function, which is 4x - 3. Therefore, G'(x) = $e^{2x² - 3x}$ · (4x - 3) = (4x - 3) $e^{2x² - 3x}$ . Choice A incorrectly provides just $e^{2x² - 3x}$ , omitting the crucial multiplication by the derivative of the exponent. Remember that when differentiating $e^{f(x)}$ , the result is always $e^{f(x)}$ · f'(x).
A force is given by $F(x)=\ln\big(5x^2-4x+1\big)$ . What is $F'(x)$ ?
Answer: $\dfrac{10x-4}{5x^2-4x+1}$ This problem requires the chain rule to differentiate a composite function. The outer function is the natural logarithm, and the inner function is 5x² - 4x + 1. To differentiate, take the derivative of the outer function, which is 1 over the inner function. Then multiply by the derivative of the inner function, which is 10x - 4, resulting in (10x - 4) / (5x² - 4x + 1). A tempting distractor is choice A, which incorrectly places the derivative in the numerator without the logarithmic structure. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a polynomial inside a logarithmic function.
A medication level is $M(t)=\sin\big(2t^3-t\big)$ ; what is $M'(t)$ ?
Answer: $\cos\big(2t^3-t\big)\big(6t^2-1\big)$ This problem requires the chain rule to differentiate a composite function. The outer function is the sine function, while the inner function is the cubic 2t³ - t. To differentiate, first take the derivative of the outer function, which is cosine of the inner function. Then, multiply by the derivative of the inner function, which is 6t² - 1, yielding cos(2t³ - t)(6t² - 1). A tempting distractor like choice E swaps cosine for sine, which is the derivative of cosine instead of sine. Always look for compositions where a function is plugged into another, such as a polynomial inside a trigonometric function, to recognize when to apply the chain rule.
A density is $\rho(x)=\dfrac{1}{\sqrt{2x+9}}$ . What is $\rho'(x)$ ?
Answer: $-\dfrac{1}{(2x+9)^{3/2}}$ This problem requires the chain rule to differentiate a composite function. The outer function is raising to the -1/2 power (since 1/sqrt is $( )^{-1/2}$ ), and the inner function is 2x + 9. To differentiate, take the derivative of the outer function, which is (-1/2) times the inner function raised to the -3/2 power. Then multiply by the derivative of the inner function, which is 2, resulting in (-1/2) $(2x + 9)^{-3/2}$ * 2 = - $(2x + 9)^{-3/2}$ , or -1 / $(2x + 9)^{3/2}$ . A tempting distractor is choice B, which incorrectly keeps an extra 1/2 factor without simplifying properly. To recognize when to use the chain rule, look for compositions where one function is substituted into another, such as a linear expression inside a root or fractional power.
A sensor's temperature is modeled by $T(t)=\big(3t^2-5t+1\big)^4$ ; what is $T'(t)$ ?
Answer: $4\big(3t^2-5t+1\big)^3\big(6t-5\big)$ This problem requires the chain rule to differentiate a composite function. The outer function is raising to the fourth power, while the inner function is the quadratic 3t² - 5t + 1. To differentiate, first take the derivative of the outer function, which is 4 times the inner function raised to the third power. Then, multiply by the derivative of the inner function, which is 6t - 5, resulting in 4(3t² - 5t + 1)³(6t - 5). A tempting distractor like choice A forgets to multiply by the inner derivative, leading to an incomplete answer. Always look for compositions where a function is plugged into another, such as a polynomial inside a power, to recognize when to apply the chain rule.
A function is $h(x)=\big(7-2x\big)^{1/3}$ . What is $h'(x)$ ?
Answer: $-\dfrac{2}{3}(7-2x)^{-2/3}$ This problem requires the chain rule to differentiate h(x) = $(7 - 2x)^{1/3}$ . The outer function is $u^{1/3}$ and the inner function is u = 7 - 2x. Applying the chain rule, we differentiate the outer function to get (1/3) $u^{-2/3}$ , then multiply by the derivative of the inner function, which is -2. Therefore, h'(x) = (1/3) $(7 - 2x)^{-2/3}$ · (-2) = -2/3 $(7 - 2x)^{-2/3}$ . Choice A incorrectly omits the derivative of the inner function (-2), providing only the partial result from differentiating the outer function. Remember that the chain rule always requires the product of the outer derivative and the inner derivative.
A cost function is $K(x)=\ln\big(7-3x^2\big)$ ; what is $K'(x)$ ?
Answer: $\frac{-6x}{7-3x^2}$ To find $K'(x)$ for $K(x)=\ln(7-3x^2)$ , we use the chain rule. The outer function is $\ln(u)$ with derivative $\frac{1}{u}$ , and the inner function is $u=7-3x^2$ with derivative $-6x$ . Applying the chain rule gives $K'(x)=\frac{1}{7-3x^2}\cdot(-6x)=\frac{-6x}{7-3x^2}$ . Choice A gives only the derivative of the outer function without the chain rule. Remember that $\frac{d}{dx}[\ln(f(x))]=\frac{f'(x)}{f(x)}$ —the derivative of the inner function goes in the numerator.
A population model is $P(t)=\big(1+e^{-2t}\big)^{-1}$ . What is $P'(t)$ ?
Answer: $\dfrac{2e^{-2t}}{\big(1+e^{-2t}\big)^2}$ This problem requires the chain rule to differentiate a function with a negative exponent. The function P(t) = $(1 + e^{-2t})^{-1}$ has an outer function f(u) = $u^{-1}$ and an inner function u = 1 + $e^{-2t}$ . By the chain rule, P'(t) = f'(u) · u'(t) = - $u^{-2}$ · (- $2e^{-2t}$ ) = $2e^{-2t}$ /(1 + $e^{-2t}$ )². Answer D incorrectly omits the $e^{-2t}$ factor from the derivative of the inner function, giving -2/(1 + $e^{-2t}$ )². To spot chain rule problems, look for any composition of functions—here, a power function applied to an expression containing an exponential.
A tank's temperature is modeled by $T(t)=\big(3t^2-5t+1\big)^4$ ; what is $T'(t)$ ?
Answer: $4(3t^2-5t+1)^3(6t-5)$ This problem requires the chain rule to differentiate $T(t)=(3t^2-5t+1)^4$ . The outer function is $u^4$ and the inner function is $u=3t^2-5t+1$ . Using the chain rule, we differentiate the outer function to get $4u^3$ , then multiply by the derivative of the inner function, which is $6t-5$ . This gives us $T'(t)=4(3t^2-5t+1)^3(6t-5)$ . Choice B incorrectly gives only the derivative of the inner function without applying the chain rule. When you see a composite function like $(expression)^n$ , always identify the inner expression first, then multiply the power rule result by the inner derivative.
A concentration is $c(t)=\ln\big(t^2+6t+10\big)$ . What is $c'(t)$ ?
Answer: $\dfrac{2t+6}{t^2+6t+10}$ This problem requires the chain rule to differentiate c(t) = ln(t² + 6t + 10). The outer function is ln(u) and the inner function is u = t² + 6t + 10. By the chain rule, we differentiate the outer function to get 1/u, then multiply by the derivative of the inner function, which is 2t + 6. Therefore, c'(t) = (1/(t² + 6t + 10)) · (2t + 6) = (2t + 6)/(t² + 6t + 10). Choice A incorrectly provides only 1/(t² + 6t + 10), forgetting to multiply by the derivative of the inner function. When differentiating ln(f(x)), always remember the formula: d/dx[ln(f(x))] = f'(x)/f(x).
A radar distance is $d(t)=\big(\frac{4}{t+1}\big)^2$ ; what is $d'(t)$ ?
Answer: $-\frac{32}{(t+1)^3}$ To differentiate $d(t)=(\frac{4}{t+1})^2=\frac{16}{(t+1)^2}$ , we can use the chain rule. The outer function is $u^2$ and inner is $u=\frac{4}{t+1}=4(t+1)^{-1}$ . The derivative of the inner is $-4(t+1)^{-2}$ . Applying chain rule: $d'(t)=2(\frac{4}{t+1})\cdot(-4(t+1)^{-2})=\frac{8}{t+1}\cdot\frac{-4}{(t+1)^2}=-\frac{32}{(t+1)^3}$ . Choice B has the wrong sign. When differentiating quotients raised to powers, watch the signs carefully—negative exponents introduce negative derivatives.
A growth factor is $G(x)=\ln\!\left(x^2+6x+10\right)$ . What is $G'(x)$ ?
Answer: $\dfrac{2x+6}{x^2+6x+10}$ This problem requires the chain rule to differentiate G(x) = ln(x² + 6x + 10). The chain rule states that when differentiating a composite function, we multiply the derivative of the outer function by the derivative of the inner function. The outer function is ln(u) with derivative 1/u, and the inner function is u = x² + 6x + 10 with derivative 2x + 6. Applying the chain rule: G'(x) = (1/(x² + 6x + 10)) · (2x + 6) = (2x + 6)/(x² + 6x + 10). Choice A incorrectly gives 1/(x² + 6x + 10), omitting the derivative of the inner function. Remember that d/dx[ln(f(x))] = f'(x)/f(x), which is a direct application of the chain rule to logarithmic functions.

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