Chain Rule - AP Calculus AB

Let $z(x)=\sqrt{\sin(x^2+1)}$ . What is $z'(x)$ ?
Answer: $\dfrac{2x\cos(x^2+1)}{2\sqrt{\sin(x^2+1)}}$ The chain rule is crucial for differentiating $z(x) = \sqrt{\sin(x^2 + 1)}$ , as it involves a composition of functions: the square root as the outer function applied to the sine, which itself is applied to the inner polynomial $x^2 + 1$ . Identify the outermost function as the square root, with derivative $\frac{1}{2\sqrt{u}}$ where $u = \sin(v)$ and $v = x^2 + 1$ . Then, multiply by the derivative of u, which is $\cos(v)$ times the derivative of v, which is $2x$ . This yields $\frac{1}{2\sqrt{\sin(x^2 + 1)}} \cdot \cos(x^2 + 1) \cdot 2x$ , simplifying to $\frac{x \cos(x^2 + 1)}{\sqrt{\sin(x^2 + 1)}}$ . A common omission is forgetting the derivative of the innermost function, such as the $2x$ from v'. The marked answer B is correct as it matches this after simplification. For pattern recognition, spot compositions where trigonometric functions enclose polynomials, indicating multiple chain rule applications.
A position is $x(t)=\cos\!\left((2t-1)^4\right)$ . What is $x'(t)$ ?
Answer: $-8(2t-1)^3\sin\!\left((2t-1)^4\right)$ For $x(t) = \cos((2t - 1)^4)$ , we have cosine of a power of a linear function. The outer function is $\cos(u)$ with derivative $-\sin(u)$ . The middle function is $u = v^4$ with derivative $4v^3$ . The inner function is $v = 2t - 1$ with derivative $2$ . Applying the chain rule through all three layers: $x'(t) = -\sin((2t - 1)^4) \cdot 4(2t - 1)^3 \cdot 2 = -8(2t - 1)^3\sin((2t - 1)^4)$ . Common errors include forgetting the negative sign from cosine's derivative or missing the factor of 2 from the innermost derivative. When differentiating nested compositions, work systematically from outside to inside, keeping track of all factors and signs.
Let $f(x)=\ln(2x^3-x)$ . What is $f'(x)$ ?
Answer: $\dfrac{6x^2-1}{2x^3-x}$ For f(x) = ln(2x³ - x), we recognize the chain rule structure: natural log (outer) of a polynomial (inner). The derivative of ln(u) is 1/u, giving us 1/(2x³ - x) for the outer derivative. We must multiply this by the derivative of the inner function (2x³ - x), which is 6x² - 1. Combining these yields (6x² - 1)/(2x³ - x). Students often forget the inner derivative and write just 1/(2x³ - x), missing the crucial multiplication step. When differentiating logarithms of expressions, always ask yourself: 'What's inside the log, and what's its derivative?' This two-step process ensures you capture both parts of the chain rule.
At time $t$ , a signal is $S(t)=\sin\!\left(4t^2-3t\right)$ . What is $S'(t)$ ?
Answer: $(8t-3)\cos(4t^2-3t)$ To differentiate $S(t)=\sin(4t^2-3t)$ , we apply the chain rule with the outer function being sine and the inner function being $4t^2-3t$ . The derivative of sine is cosine, so we get $\cos(4t^2-3t)$ for the outer derivative. We must multiply this by the derivative of the inner function: $(4t^2-3t)'=8t-3$ . Therefore, $S'(t)=(8t-3)\cos(4t^2-3t)$ . A common mistake is to stop after differentiating just the outer function, giving $\cos(4t^2-3t)$ alone, which misses the chain rule multiplication. Remember: when differentiating a trig function with a non-simple argument, always multiply by the derivative of what's inside the trig function.
Let $g(x)=\left(2x^4-x\right)^5$ . What is $g'(x)$ ?
Answer: $5(2x^4-x)^4(8x^3-1)$ To differentiate $g(x) = (2x^4 - x)^5$ , we apply the chain rule with the power function as the outer function and the polynomial as the inner function. The outer function is $f(u) = u^5$ with derivative $f'(u) = 5u^4$ . The inner function is $u = 2x^4 - x$ with derivative $u' = 8x^3 - 1$ . By the chain rule: $g'(x) = 5(2x^4 - x)^4 \cdot (8x^3 - 1) = 5(2x^4 - x)^4(8x^3 - 1)$ . A common mistake is to only apply the power rule without considering the inner function (answer A) or to incorrectly compute the inner derivative. Remember the pattern: when differentiating $(\text{expression})^n$ , the result is $n(\text{expression})^{n-1} \cdot (\text{derivative of expression})$ .
A revenue function is $R(x)=\sqrt[3]{x^2+4x}$ . What is $R'(x)$ ?
Answer: $\dfrac{2x+4}{3}(x^2+4x)^{-2/3}$ R(x) = ∛(x² + 4x) is $(u)^{1/3}$ with u(x) = x² + 4x, so chain rule gives (1/3) $u^{-2/3}$ · u' = (1/3) $(x² + 4x)^{-2/3}$ · (2x + 4). Simplifying, it's (2x + 4)/3 · $(x² + 4x)^{-2/3}$ . The fractional power is key. Common omission: Forgetting the (2x + 4) factor. Some mishandle the exponent. Pattern: Look for roots of polynomials and chain their derivatives.
For $\theta(t)=\dfrac{1}{\sqrt{\sin t}}$ , what is $\theta'(t)$ ?
Answer: $ -\dfrac{\cos t}{2}(\sin t)^{-3/2} $ $\theta(t) = \dfrac{1}{\sqrt{\sin t}} = (\sin t)^{-1/2}$ . Derivative: $-\dfrac{1}{2} (\sin t)^{-3/2} \cos t$ . Outer: Negative power of sin. Common omission: Sign or exponent error. Matches choice B. Verified. Recognize negative roots as powers for chain rule application.
For $M(x)=\sin(\pi x^3)$ , what is $M'(x)$ ?
Answer: $3\pi x^2\cos(\pi x^3)$ M(x) = $\sin(\pi x^3)$ requires the chain rule due to the composition: sine (outer) of $\pi$ times $x^3$ (inner). The derivative is $\cos(\pi x^3)$ multiplied by the derivative of the inside, which is $\pi \cdot 3x^2$ . So, $M'(x) = 3\pi x^2 \cos(\pi x^3)$ . Students often omit the constant $\pi$ or the $3x^2$ from the power rule. This matches choice B. Verification shows the marked answer is accurate. Look for trig functions with polynomial arguments raised to powers, cueing chain rule with power rule inside.
A function is $f(x)=\sqrt[5]{\sin(3x+1)}$ . What is $f'(x)$ ?
Answer: $\dfrac{3}{5}(\sin(3x+1))^{-4/5}\cos(3x+1)$ $f(x) = [\sin(3x + 1)]^{1/5}$ is fractional power (outer) of sine (inner) of linear. Derivative: $\frac{1}{5} [\sin w]^{-4/5} \cos w \cdot 3 = \frac{3}{5} \cos(3x + 1) [\sin(3x + 1)]^{-4/5}$ . Outer-inner clear. Omission: Missing the 3. Choice B matches. Verified correct. Pattern: Fractional powers on trig signal chain rule for growth models.
A chemical concentration is $C(t)=\dfrac{1}{(t^2+3)^4}$ . What is $C'(t)$ ?
Answer: $-8t(t^2+3)^{-5}$ C(t) = 1/ $(t² + 3)^4$ = $(t² + 3)^{-4}$ . Chain: -4 $(t² + 3)^{-5}$ · 2t = -8t $(t² + 3)^{-5}$ . Negative power and inner quadratic. Omission: Missing the 2t or sign. Some keep -4 only. Pattern: Reciprocals of powers chain with negative exponents.
For $B(t)=\cos^3(t^2)$ , what is $B'(t)$ ?
Answer: $-6t\sin(t^2)\cos^2(t^2)$ B(t) = $\cos^3$ ( $t^2$ ) is [cos(u)]^3 where u = $t^2$ , so chain rule for power (outer) and trig (inner). Derivative: 3 $\cos^2$ (u) (-sin(u)) 2t = -6t sin( $t^2$ ) $\cos^2$ ( $t^2$ ). Outer-inner: Power on trig of quadratic. Omission: Missing the 2t or the negative sign. Matches choice C. Confirmed correct. Pattern: Powers on trig functions of polynomials need extended chain rule.
Let $m(x)=\sin\!\left(\dfrac{1}{x^2+3}\right)$ . What is $m'(x)$ ?
Answer: $\dfrac{-2x\cos\!\left(\dfrac{1}{x^2+3}\right)}{(x^2+3)^2}$ For $m(x) = \sin\left(\frac{1}{x^2 + 3}\right)$ , we have sine of a reciprocal function. The outer function is $\sin(u)$ with derivative $\cos(u)$ . The inner function is $u = \frac{1}{x^2 + 3} = (x^2 + 3)^{-1}$ with derivative $u' = -(x^2 + 3)^{-2} \cdot 2x = \frac{-2x}{(x^2 + 3)^2}$ (using the chain rule again). Therefore: $m'(x) = \cos\left(\frac{1}{x^2 + 3}\right) \cdot \frac{-2x}{(x^2 + 3)^2} = \frac{-2x\cos\left(\frac{1}{x^2 + 3}\right)}{(x^2 + 3)^2}$ . A common error is forgetting the negative sign when differentiating the reciprocal. When you see a function of a reciprocal, remember that $\frac{d}{dx}\left[\frac{1}{g(x)}\right] = \frac{-g'(x)}{[g(x)]^2}$ .
A trajectory is $y(x)=\left(3x^2-4x+1\right)^5$ . What is $y'(x)$ ?
Answer: $5(6x-4)(3x^2-4x+1)^4$ For $y(x)=(3x^2-4x+1)^5$ , we have a power function as the outer function and $3x^2-4x+1$ as the inner function. Using the power rule on the outer function gives us $5(3x^2-4x+1)^4$ . The inner function's derivative is $(3x^2-4x+1)'=6x-4$ . By the chain rule, we multiply: $y'(x)=5(3x^2-4x+1)^4\cdot(6x-4)=5(6x-4)(3x^2-4x+1)^4$ . A common mistake is to multiply the exponent by each term inside the parentheses, yielding something like $(15x^2-20x+5)(3x^2-4x+1)^4$ , which confuses the power rule with distribution. Remember: when differentiating $(\text{stuff})^n$ , you get $n(\text{stuff})^{n-1}\cdot(\text{derivative of stuff})$ .
If $y(x)=\ln\!\left(\sqrt{1-6x}\right)$ , what is $\dfrac{dy}{dx}$ ?
Answer: $\dfrac{-3}{1-6x}$ To differentiate $y(x) = \ln(\sqrt{1 - 6x})$ , we can first simplify using logarithm properties: $\ln(\sqrt{1 - 6x}) = \ln((1 - 6x)^{1/2}) = \frac{1}{2}\ln(1 - 6x)$ . Now differentiating: $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1 - 6x} \cdot (-6) = \frac{-6}{2(1 - 6x)} = \frac{-3}{1 - 6x}$ . Alternatively, using the chain rule directly: outer function is $\ln(u)$ with derivative $\frac{1}{u}$ , middle function is $\sqrt{v}$ with derivative $\frac{1}{2\sqrt{v}}$ , and inner function is $v = 1 - 6x$ with derivative $-6$ . This gives $\frac{1}{\sqrt{1 - 6x}} \cdot \frac{1}{2\sqrt{1 - 6x}} \cdot (-6) = \frac{-3}{1 - 6x}$ . Remember: simplifying logarithms before differentiating often makes the calculation easier.
Let $S(x)=\sqrt{1+\tan(3x)}$ . What is $S'(x)$ ?
Answer: $\dfrac{3\sec^2(3x)}{2\sqrt{1+\tan(3x)}}$ S(x) = √(1 + tan(3x)) is root outer of 1 + tan inner of linear. Derivative: (1/(2√(1 + tan(3x)))) * $\sec^2$ (3x) * 3. Structure evident. Omission: Forgetting $\sec^2$ or 3. Matches choice C. Confirmed. Pattern: Roots of tan derivatives use chain with $\sec^2$ .
For $\phi(x)=\sqrt{\dfrac{1}{1-4x}}$ , what is $\phi'(x)$ ?
Answer: $\dfrac{2}{(1-4x)^2\sqrt{1/(1-4x)}}$ ϕ(x) = √[1/(1-4x)] is root of reciprocal. Derivative simplifies to 2 $(1-4x)^{-3/2}$ . Outer-inner: Root of inverse. Common omission: Algebra errors in exponents. Choice C matches. Verified. Strategy: Reciprocals inside roots need careful chain and simplification.
A temperature model is $T(x)=\cos\!\left((x^2+1)^3\right)$ . What is $T'(x)$ ?
Answer: $-\sin\!\left((x^2+1)^3\right)\cdot 6x(x^2+1)^2$ For $T(x)=\cos((x^2+1)^3)$ , we have a composition of three functions: cosine (outermost), cube (middle), and $x^2+1$ (innermost). Working from outside to inside: the derivative of $\cos(u)$ is $-\sin(u)$ , giving us $-\sin((x^2+1)^3)$ . Next, we need the derivative of $(x^2+1)^3$ , which by the power rule is $3(x^2+1)^2$ times the derivative of $x^2+1$ , which is $2x$ . So the full inner derivative is $3(x^2+1)^2\cdot 2x=6x(x^2+1)^2$ . Therefore, $T'(x)=-\sin((x^2+1)^3)\cdot 6x(x^2+1)^2$ . A common error is stopping after one application of the chain rule, missing the nested composition. When you see multiple levels of composition, work systematically from outside to inside, multiplying all the derivative factors together.
Let $g(x)=\cos(5x^2)$ . What is $g'(x)$ ?
Answer: $-10x\sin(5x^2)$ g(x) = cos(5x²) uses the chain rule with outer cos(u) and inner u(x) = 5x². The derivative is -sin(u) · u' = -sin(5x²) · 10x = -10x sin(5x²). Don't forget the negative from cos derivative. A common omission is missing the 10x, treating it as cos(5x) instead. Some forget the chain entirely. To recognize patterns, note trig functions of quadratics demand chaining the polynomial derivative.
Let $W(x)=\big(5-\sin x\big)^{10}$ . What is $W'(x)$ ?
Answer: $-10(5-\sin x)^9\cos x$ W(x) = (5 - sin x)^{10} is power outer of 5 - sin inner. Derivative: 10(5 - sin x)^9 * (-cos x). Outer-inner clear. Common omission: Missing negative from sin derivative. Choice C matches. Verified. Strategy: Powers of trig subtractions use chain with signs.
For $f(x)=e^{\sin(3x)}$ , what is $f'(x)$ ?
Answer: $3e^{\sin(3x)}\cos(3x)$ For f(x) = $e^{\sin(3x)}$ , the chain rule applies to the exponential outer function with inner u(x) = $\sin(3x)$ . The derivative is $e^u$ times u'(x), where u' = $\cos(3x) \cdot 3 = 3 \cos(3x)$ , so f'(x) = $3 e^{\sin(3x)} \cos(3x)$ . Note the additional chain for sin(3x) itself. A common omission is forgetting the 3 from the inner chain of 3x. Students might drop the cos term entirely. Recognize chain rule needs when exponentials contain trig functions of linear multiples, always differentiating layer by layer.

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