Chain Rule — MH-CET Basic Maths MCQs with Solutions
Free MH-CET Basic Maths Chain Rule MCQs with step-by-step solutions (30 questions). Part of Differentiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Chain Rule · easy · theory
The chain rule for differentiating a composite function y = f(u(x)) states:
A. $\dfrac{dy}{dx} = \dfrac{dy}{du} \div \dfrac{du}{dx}$
B. $\dfrac{dy}{dx} = \dfrac{du}{dx}$ only
C. $\dfrac{dy}{dx} = \dfrac{dy}{du} + \dfrac{du}{dx}$
D. $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ ✓ Correct
Solution: Chain rule: differentiate the outer function with respect to the inner, then multiply by the derivative of the inner: dy/dx = (dy/du)(du/dx).
Q2 — Chain Rule · easy · theory
For y = sin(x²), the correct identification of the outer and inner functions is:
A. Outer f(u) = u², inner u(x) = sin x
B. No composite structure exists
C. Outer f(u) = sin u, inner u(x) = x² ✓ Correct
D. Both are sin x
Solution: The last operation applied is sine (of x²), so sin is the outer function and u = x² is the inner one.
Q3 — Chain Rule · easy · theory
For y = sin²x (i.e. (sin x)²), the outer and inner functions are:
A. Outer f(u) = u², inner u(x) = sin x ✓ Correct
B. It is not a composite function
C. Outer f(u) = 2u, inner u(x) = sin x
D. Outer f(u) = sin u, inner u(x) = x²
Solution: sin²x means (sin x)² — the squaring is applied last, so the square is the outer function. Pitfall: sin²x and sin(x²) have opposite structures!
Q4 — Chain Rule · easy · theory
The derivative of e^{u(x)} with respect to x is:
A. $u'(x)\,e^{u'(x)}$
B. $e^{u}\,u'(x)$ ✓ Correct
C. $u\,e^{u-1}\,u'(x)$
D. $e^{u}$ only
Solution: d/dx e^u = e^u · du/dx. The exponential reproduces itself; the chain rule appends the inner derivative.
Q5 — Chain Rule · easy · theory
The derivative of ln[u(x)] with respect to x is:
A. $\ln u \cdot u'(x)$
B. $\dfrac{1}{u(x)}$ only
C. $\dfrac{u(x)}{u'(x)}$
D. $\dfrac{u'(x)}{u(x)}$ ✓ Correct
Solution: d/dx ln u = (1/u)·u' = u'/u — the inner derivative must multiply 1/u.
Q6 — Chain Rule · easy · theory
The general power rule (chain-rule form) for y = [u(x)]ⁿ gives dy/dx =
A. $[u'(x)]^n$
B. $n\,u'(x)^{n-1}$
C. $n[u(x)]^{n-1}$ only
D. $n[u(x)]^{n-1}\,u'(x)$ ✓ Correct
Solution: Power rule + chain rule: bring the power down, reduce it by one, multiply by the inner derivative u'.
Q7 — Chain Rule · medium · theory
A function like $y = e^{\sqrt{\tan x}}$ requires the chain rule applied:
A. Four times
B. Not at all
C. Once only
D. Twice (three nested layers: exponential, square root, tangent) ✓ Correct
Solution: Layers: e^( ), √( ), tan( ). Peeling two links: dy/dx = e^{√tan x} · (1/(2√tan x)) · sec²x — the chain rule chains twice.
Q8 — Chain Rule · easy · theory
Which of the following functions does NOT need the chain rule to differentiate?
A. $y = x^3 + 5x$ ✓ Correct
B. $y = \cos(3x)$
C. $y = e^{2x}$
D. $y = \ln(x^2+1)$
Solution: x³ + 5x is a plain polynomial in x (inner function is x itself). The others are genuine composites: cos(3x), ln(x²+1), e^{2x}.
Q9 — Chain Rule · easy · theory
The derivative of sin(kx) (k constant) is k·cos(kx). The factor k appears because:
A. Sine always produces a factor k
B. The chain rule multiplies by the derivative of the inner function kx ✓ Correct
C. cos(kx) requires it dimensionally only
D. It is a convention
Solution: Inner u = kx has du/dx = k, so d/dx sin(kx) = cos(kx)·k. Forgetting the inner derivative is the most common chain-rule error.
Q10 — Chain Rule · medium · theory
If y = f(u), u = g(v) and v = h(x), then dy/dx equals:
A. $\dfrac{dy}{du}\cdot\dfrac{du}{dv}\cdot\dfrac{dv}{dx}$ ✓ Correct
B. $\dfrac{dy}{dv}\cdot\dfrac{dv}{du}$
C. $\dfrac{dy}{du}\cdot\dfrac{dv}{dx}$ only
D. $\dfrac{dy}{du}+\dfrac{du}{dv}+\dfrac{dv}{dx}$
Solution: For three layers the chain extends: dy/dx = (dy/du)(du/dv)(dv/dx) — a product of all the link derivatives.
Q11 — Chain Rule · medium · theory
The most common ERROR when differentiating cos(x³) is writing:
A. −3x² sin(x³)
B. Using the minus sign with sine
C. −sin(x³) and forgetting the factor 3x² ✓ Correct
D. Keeping x³ inside the sine
Solution: Correct: d/dx cos(x³) = −sin(x³)·3x². Dropping the inner derivative 3x² is the classic mistake the chain rule guards against.
Q12 — Chain Rule · easy · theory
The derivative of $y = \sqrt{u(x)}$ by the chain rule is:
A. $\dfrac{1}{2\sqrt{u(x)}}$ only
B. $\dfrac{2u'(x)}{\sqrt{u(x)}}$
C. $\sqrt{u'(x)}$
D. $\dfrac{u'(x)}{2\sqrt{u(x)}}$ ✓ Correct
Solution: √u = u^{1/2} ⇒ derivative = (1/2)u^{−1/2}·u' = u'/(2√u).
Q13 — Chain Rule · easy · numerical
If $y = \sin(x^2)$, then $\dfrac{dy}{dx}$ =
A. $2x\cos(x^2)$ ✓ Correct
B. $-2x\cos(x^2)$
C. $\cos(x^2)$
D. $2x\sin(x^2)$
Solution: Outer sin → cos(x²); inner x² → 2x. dy/dx = cos(x²)·2x = 2x cos(x²).
Q14 — Chain Rule · easy · numerical
If $y = e^{3x+1}$, then $\dfrac{dy}{dx}$ =
A. $3e^{3x+1}$ ✓ Correct
B. $e^{3x+1}$
C. $3e^{3x}$
D. $(3x+1)e^{3x}$
Solution: d/dx e^u = e^u u′ with u = 3x + 1, u′ = 3 ⇒ 3e^{3x+1}.
Q15 — Chain Rule · easy · numerical
If $y = \ln(\cos x)$, then $\dfrac{dy}{dx}$ =
A. $\tan x$
B. $-\tan x$ ✓ Correct
C. $\dfrac{1}{\cos x}$
D. $-\cot x$
Solution: u = cos x: dy/dx = u′/u = (−sin x)/cos x = −tan x.
Q16 — Chain Rule · easy · numerical
If $y = \sqrt{x^2+4}$, then $\dfrac{dy}{dx}$ =
A. $\sqrt{2x}$
B. $\dfrac{x}{\sqrt{x^2+4}}$ ✓ Correct
C. $\dfrac{2x}{\sqrt{x^2+4}}$
D. $\dfrac{1}{2\sqrt{x^2+4}}$
Solution: dy/dx = (2x)/(2√(x²+4)) = x/√(x²+4).
Q17 — Chain Rule · easy · numerical
If $y = \cos(5x)$, then $\dfrac{dy}{dx}$ =
A. $-5\cos(5x)$
B. $-\sin(5x)$
C. $5\sin(5x)$
D. $-5\sin(5x)$ ✓ Correct
Solution: d/dx cos u = −sin u · u′; u′ = 5 ⇒ −5 sin(5x).
Q18 — Chain Rule · easy · numerical
If $y = \tan(2x)$, then $\dfrac{dy}{dx}$ =
A. $2\sec^2(2x)$ ✓ Correct
B. $\sec^2(2x)$
C. $-2\sec^2(2x)$
D. $2\tan(2x)\sec(2x)$
Solution: d/dx tan u = sec²u · u′ = sec²(2x)·2.
Q19 — Chain Rule · easy · numerical
If $y = (3x^2+1)^4$, then $\dfrac{dy}{dx}$ =
A. $24x(3x^2+1)^3$ ✓ Correct
B. $12x(3x^2+1)^3$
C. $6x(3x^2+1)^4$
D. $4(3x^2+1)^3$
Solution: Power rule: 4(3x²+1)³ × (6x) = 24x(3x²+1)³.
Q20 — Chain Rule · easy · numerical
If $y = e^{x^2}$, then $\dfrac{dy}{dx}$ =
A. $2x\,e^{x^2}$ ✓ Correct
B. $e^{x^2}$
C. $x^2 e^{x^2-1}$
D. $2e^{x}$
Solution: e^u with u = x²: dy/dx = e^{x²}·2x.
Q21 — Chain Rule · easy · numerical
If $y = \ln(x^2+1)$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{2x}{x^2+1}$ ✓ Correct
B. $\dfrac{x}{x^2+1}$
C. $2x\ln(x^2+1)$
D. $\dfrac{1}{x^2+1}$
Solution: u′/u = 2x/(x²+1).
Q22 — Chain Rule · medium · numerical
If $y = \sin^3(2x)$, then $\dfrac{dy}{dx}$ =
A. $3\sin^2(2x)\cos(2x)$
B. $6\sin^2(2x)\cos(2x)$ ✓ Correct
C. $3\cos^2(2x)$
D. $6\sin^2(2x)$
Solution: Three layers: (·)³, sin(·), 2x. dy/dx = 3sin²(2x)·cos(2x)·2 = 6 sin²(2x)cos(2x).
Q23 — Chain Rule · easy · numerical
If $y = \sqrt{\sin x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{\cos x}{\sqrt{\sin x}}$
B. $\dfrac{\cos x}{2\sqrt{\sin x}}$ ✓ Correct
C. $\dfrac{1}{2\sqrt{\sin x}}$
D. $\dfrac{-\cos x}{2\sqrt{\sin x}}$
Solution: dy/dx = u′/(2√u) with u = sin x ⇒ cos x/(2√sin x).
Q24 — Chain Rule · medium · numerical
If $y = e^{\sin x}$, its derivative at $x = 0$ is:
A. −1
B. 0
C. e
D. 1 ✓ Correct
Solution: dy/dx = e^{sin x}·cos x. At x = 0: e⁰ × cos 0 = 1 × 1 = 1.
Q25 — Chain Rule · medium · numerical
If $y = \cos^2 x$, then $\dfrac{dy}{dx}$ =
A. $-\sin 2x$ ✓ Correct
B. $\sin 2x$
C. $2\cos x \sin x$
D. $-2\cos x$
Solution: dy/dx = 2cos x·(−sin x) = −2 sin x cos x = −sin 2x (double-angle simplification).
Q26 — Chain Rule · medium · numerical
If $y = \ln(\ln x)$ (x > 1), then $\dfrac{dy}{dx}$ =
A. $\dfrac{1}{\ln x}$
B. $\dfrac{x}{\ln x}$
C. $\dfrac{1}{x\ln x}$ ✓ Correct
D. $\dfrac{\ln x}{x}$
Solution: Outer ln(u), u = ln x: dy/dx = (1/ln x)·(1/x) = 1/(x ln x).
Q27 — Chain Rule · easy · numerical
If $y = (1 - 2x)^5$, then $\dfrac{dy}{dx}$ =
A. $10(1-2x)^4$
B. $-10(1-2x)^4$ ✓ Correct
C. $5(1-2x)^4$
D. $-5(1-2x)^4$
Solution: 5(1−2x)⁴ × (−2) = −10(1−2x)⁴. Pitfall: don't drop the inner derivative −2.
Q28 — Chain Rule · medium · numerical
If $y = \sec(x^2)$, then $\dfrac{dy}{dx}$ =
A. $2x\tan^2(x^2)$
B. $2x\sec^2(x^2)$
C. $2x\sec(x^2)\tan(x^2)$ ✓ Correct
D. $\sec(x^2)\tan(x^2)$
Solution: d/dx sec u = sec u tan u · u′ = sec(x²)tan(x²)·2x.
Q29 — Chain Rule · medium · numerical
The derivative of $y = \sin(3x)$ evaluated at $x = 0$ is:
A. 0
B. −3
C. 1
D. 3 ✓ Correct
Solution: dy/dx = 3cos(3x); at x = 0: 3cos0 = 3.
Q30 — Chain Rule · medium · numerical
If $y = e^{\sqrt{x}}$, then $\dfrac{dy}{dx}$ =
A. $\sqrt{x}\,e^{\sqrt{x}-1}$
B. $\dfrac{e^{\sqrt{x}}}{2\sqrt{x}}$ ✓ Correct
C. $\dfrac{e^{\sqrt{x}}}{\sqrt{x}}$
D. $e^{\sqrt{x}}$
Solution: u = √x, u′ = 1/(2√x): dy/dx = e^{√x}/(2√x).