Chain Rule
Help Questions
AP Calculus AB › Chain Rule
For $d(x)=\arcsin(2x-3)$, what is $d'(x)$?
$\dfrac{1}{\sqrt{1-(2)^2}}$
$\dfrac{1}{2\sqrt{1-(2x-3)^2}}$
$\dfrac{1}{\sqrt{1-(2x-3)^2}}$
$\dfrac{2x-3}{\sqrt{1-(2x-3)^2}}$
$\dfrac{2}{\sqrt{1-(2x-3)^2}}$
Explanation
d(x) = arcsin(2x - 3) is inverse sine outer of linear. Derivative: 1/√(1 - (2x - $3)^2$) * 2. Structure: Inverse trig of linear. Omission: Forgetting the 2. Matches choice B. Confirmed. Pattern: Inverse trig derivatives always include chain for argument.
For $F(x)=(\sin x)^5$, what is $F'(x)$?
$(\cos x)^5$
$4(\sin x)^3\cos x$
$5(\sin x)^4\cos x$
$5\sin(x^4)\cos x$
$5(\sin x)^4$
Explanation
F(x) = (sin $x)^5$ is power with inner sin x. Chain rule: 5 (sin $x)^4$ · cos x. Power reduces, multiply by cos x. Omission: Forgetting cos x. Some use 4 instead of 5. Pattern: Powers of trig functions need chaining the trig derivative.
A chemical concentration is $C(t)=\dfrac{1}{(t^2+3)^4}$. What is $C'(t)$?
$-4(t^2+3)^{-5}$
$-8t(t^2+3)^{-4}$
$\dfrac{-8}{t}(t^2+3)^{-5}$
$-8t(t^2+3)^{-5}$
$-4(t^2+3)^{-4}$
Explanation
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 $p(x)=\arctan(3x^2-1)$, what is $p'(x)$?
$\dfrac{3x^2-1}{1+(3x^2-1)^2}$
$\dfrac{1}{1+(3x^2-1)^2}$
$\dfrac{6x}{1+3x^2-1}$
$\dfrac{6x}{1+(3x^2-1)^2}$
$\dfrac{1}{1+(6x)^2}$
Explanation
p(x) = arctan(3x² - 1), outer arctan(u), inner u = 3x² - 1. Derivative: 1/(1 + u²) · 6x. Arctan derivative times inner. Omission: Missing 6x. Some square u wrong. Pattern: Inverse trig with quadratics chain the polynomial derivative.
Let $W(x)=\big(5-\sin x\big)^{10}$. What is $W'(x)$?
$-9(5-\sin x)^{10}\cos x$
$10(5-\sin x)^9$
$10(5-\sin x)^9\cos x$
$-10(5-\sin x)^9\cos x$
$-10(5-\cos x)^9\sin x$
Explanation
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.
Let $S(x)=\sqrt{1+\tan(3x)}$. What is $S'(x)$?
$\dfrac{3\tan(3x)}{2\sqrt{1+\tan(3x)}}$
$\dfrac{\sec^2(3x)}{2\sqrt{1+\tan(3x)}}$
$\dfrac{3\sec^2(3x)}{2\sqrt{1+\tan(3x)}}$
$\dfrac{1}{2\sqrt{1+\tan(3x)}}$
$\dfrac{3\sec(3x)}{2\sqrt{1+\tan(3x)}}$
Explanation
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 $sec^2$.
A wave is $w(t)=\sin^2(4t)$. What is $w'(t)$?
$2\sin(4t)\cos(4t)$
$8\sin^2(4t)$
$2\sin(4t)$
$8\sin(4t)\cos(4t)$
$4\sin(8t)$
Explanation
w(t) = sin²(4t) = [sin(4t)]². Chain: 2 sin(4t) · cos(4t) · 4 = 8 sin(4t) cos(4t). Power and inner sin(4t). Omission: Forgetting the 4. Some use sin derivative alone. Pattern: Squared trig with multiples require full chaining.
A trajectory is $y(x)=\left(3x^2-4x+1\right)^5$. What is $y'(x)$?
$5(6x-4)(3x^2-4x+1)^4$
$(15x^2-20x+5)(3x^2-4x+1)^4$
$5(3x^2-4x+1)^5(6x-4)$
$5(3x^2-4x+1)^4$
$(6x-4)(3x^2-4x+1)^5$
Explanation
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 $n(\text{stuff})^{n-1}\cdot(\text{derivative of stuff})$.
For $k(x)=\sqrt{x^3-4x}$, what is $k'(x)$?
$\dfrac{3x^2-4}{\sqrt{x^3-4x}}$
$\dfrac{3x^2-4}{2}(x^3-4x)^{2}$
$\dfrac{1}{2\sqrt{x^3-4x}}$
$\dfrac{3x^2-4}{2\sqrt{x^3}}$
$\dfrac{3x^2-4}{2\sqrt{x^3-4x}}$
Explanation
k(x) = √(x³ - 4x), outer square root $u^{1/2}$, u = x³ - 4x. Derivative: (1/2) $u^{-1/2}$ · (3x² - 4) = (3x² - 4)/(2 √(x³ - 4x)). Cubic inner. Omission: Forgetting 3x² - 4. Some drop 1/2. Pattern: Roots of cubics chain the polynomial derivative.
A position is $x(t)=\cos!\left((2t-1)^4\right)$. What is $x'(t)$?
$-8(2t-1)^3\sin!\left((2t-1)^4\right)$
$-8(2t-1)^4\sin(2t-1)$
$-\sin!\left((2t-1)^4\right)$
$-4(2t-1)^3\sin!\left((2t-1)^4\right)$
$8(2t-1)^3\sin!\left((2t-1)^4\right)$
Explanation
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.