Chain Rule
Help Questions
AP Calculus AB › Chain Rule
For $J(x)=\sec(\ln x)$, what is $J'(x)$?
$\dfrac{1}{x}\sec(\ln x)$
$\dfrac{\sec x\tan x}{x}$
$x\sec(\ln x)\tan(\ln x)$
$\dfrac{\sec(\ln x)\tan(\ln x)}{x}$
$\sec(\ln x)\tan(\ln x)$
Explanation
J(x) = sec(ln x), outer sec(u), u = ln x. Derivative: sec(u) tan(u) · (1/x). Log inner. Omission: Missing 1/x. Some forget tan. Pattern: Trig of logs chain the 1/x derivative.
Let $v(x)=e^{x\cos x}$. What is $v'(x)$?
$e^{x\cos x}(\cos x)$
$e^{x\cos x}(x\cos x)$
$e^{x\cos x}(\cos x- x\sin x)$
$e^{x\cos x}(\sin x+ x\cos x)$
$e^{x\sin x}(\cos x- x\sin x)$
Explanation
v(x) = $e^{x cos x}$, outer $e^u$, u = x cos x. u' needs product rule: cos x - x sin x. Derivative: $e^u$ · (cos x - x sin x). Composite exponent. Omission: Missing product rule in u'. Some simplify wrong. Pattern: Exponentials with products chain including product derivative.
Let $A(x)=\ln!\big((x-2)^2+5\big)$. What is $A'(x)$?
$\dfrac{(x-2)^2+5}{2(x-2)}$
$\dfrac{2x}{(x-2)^2+5}$
$\dfrac{2(x-2)}{(x-2)^2+5}$
$2(x-2)\ln((x-2)^2+5)$
$\dfrac{1}{(x-2)^2+5}$
Explanation
A(x) = $\ln((x - 2)^2 + 5)$, outer $\ln(u)$, $u = (x - 2)^2 + 5$. Derivative: $\frac{1}{u} \cdot 2(x - 2)$. Inner quadratic. Omission: Missing $2(x - 2)$. Some treat as $\ln(x)$. Pattern: Logs of quadratics chain the quadratic derivative.
A function is $Z(t)=\ln!\big(\sqrt{t^4+9}\big)$. What is $Z'(t)$?
$\dfrac{2t^3}{\sqrt{t^4+9}}$
$\dfrac{4t^3}{t^4+9}$
$\dfrac{2t^3}{t^4+9}$
$\dfrac{1}{\sqrt{t^4+9}}$
$\dfrac{4t^3}{\sqrt{t^4+9}}$
Explanation
Z(t) = $ln(√(t^4$ + 9)) simplifies to (1/2) $ln(t^4$ + 9), chain rule: (1/2) * $(1/(t^4$ + 9)) * $4t^3$ = $2t^3$$/(t^4$ + 9). Structure: Log of root. Omission: Forgetting the 1/2 factor. Matches choice B. Correct. Pattern: Logs of roots simplify but still need chain.
A cost is $K(x)=(\ln x)^4$. What is $K'(x)$?
$4(\ln x)^3$
$\dfrac{4(\ln x)^3}{x}$
$\dfrac{4}{\ln x}$
$4x^3\ln x$
$\dfrac{(\ln x)^4}{x}$
Explanation
To find $K'(x)$ for $K(x) = (\ln x)^4$, apply the chain rule since this is a power function (outer) composed with the natural logarithm (inner). The outer function is $u^4$ with derivative $4u^3$, where $u = \ln x$, and the inner derivative is $\frac{1}{x}$. Thus, $K'(x) = 4(\ln x)^3 \cdot \frac{1}{x}$. A common omission is neglecting the inner derivative $1/x$, leading to just $4(\ln x)^3$. This matches choice B directly. Independently verifying, the calculation confirms the marked answer. Recognize this pattern in powers of logarithms, always multiplying by the inner function's derivative.
A decay model is $D(t)=e^{-2t^3+1}$. What is $D'(t)$?
$(-2t^3+1)e^{-2t^3+1}$
$(-6t^2)e^{-2t^3+1}$
$(-6t^2)e^{2t^3-1}$
$-6t^2+e^{-2t^3+1}$
$e^{-2t^3+1}$
Explanation
D(t) = $e^{-2t³ + 1}$ is exponential with inner u(t) = -2t³ + 1. Derivative: $e^u$ · u' = $e^{-2t³ + 1}$ · (-6t²). The cubic requires careful differentiation. Common omission: Forgetting the -6t². Some flip the sign. Pattern recognition: Exponentials with polynomials inside signal chain rule, multiply by polynomial derivative.
A current is $I(t)=\cos(e^{t})$. What is $I'(t)$?
$e^{t}\cos(e^{t})$
$-\sin(e^{t})\cos(e^{t})$
$-\sin(e^{t})$
$-e^{t}\sin(e^{t})$
$-\sin(t)e^{t}$
Explanation
I(t) = $cos(e^t$), outer cos(u), u = $e^t$. Derivative: -sin(u) · $e^t$ = $-e^t$ $sin(e^t$). Exponential inner. Omission: Forgetting $e^t$. Some miss negative. Pattern: Trig of exponentials chain the exponential derivative.
A revenue function is $R(x)=\sqrt3{x^2+4x}$. What is $R'(x)$?
$\dfrac{2}{3}(x+2)(x^2+4x)^{-1/3}$
$\dfrac{2x+4}{3}(x^2+4x)^{2/3}$
$\dfrac{1}{3}(x^2+4x)^{-2/3}$
$(2x+4)(x^2+4x)^{-2/3}$
$\dfrac{2x+4}{3}(x^2+4x)^{-2/3}$
Explanation
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.
A path is $y(x)=\ln!\big(\sqrt{1-3x}\big)$. What is $y'(x)$?
$\dfrac{1}{\sqrt{1-3x}}$
$\dfrac{-3}{\sqrt{1-3x}}$
$\dfrac{-3}{2(1-3x)}$
$\dfrac{-3}{2\sqrt{1-3x}}$
$\dfrac{1}{2(1-3x)}$
Explanation
y(x) = ln(√(1 - 3x)) simplifies to (1/2) ln(1 - 3x), so chain rule on log: (1/2) · 1/(1 - 3x) · (-3) = -3/(2(1 - 3x)). Outer log, inner sqrt then linear. Common omission: Forgetting the 1/2 from sqrt or the -3. Some don't simplify first. Pattern: Logs of roots suggest rewriting for easier chaining.
For $B(t)=\cos^3(t^2)$, what is $B'(t)$?
$-3\sin(t)\cos^2(t^2)$
$-3\sin(t^2)\cos^2(t^2)$
$-6t\cos^2(t^2)$
$-6t\sin(t^2)\cos^2(t^2)$
$3\cos^2(t^2)$
Explanation
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.