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Differentiation — MH-CET Basic Maths MCQs with Solutions
Free MH-CET Basic Maths Differentiation MCQs with step-by-step solutions covering Chain Rule, Product Rule, Quotient Rule. Practise online on Prepizo — no login needed.
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Sample 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 · 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}.
Q8 — 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.
Q9 — 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).
Q10 — 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).
Q11 — 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}.
Q12 — 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.
Q13 — 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).
Q14 — 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).
Q15 — 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.
Q16 — 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.
Q17 — 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.
Q18 — 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).
Q19 — 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).
Q20 — 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.
Q21 — Product Rule · easy · theory
The product rule for y = u(x)·v(x) states dy/dx =
A. $u\,v' - v\,u'$
B. $u\,v' + v\,u'$ ✓ Correct
C. $\dfrac{u'v - uv'}{v^2}$
D. $u'\,v'$
Solution: Product rule: (uv)′ = u v′ + v u′ — "first times derivative of second, plus second times derivative of first."
Q22 — Product Rule · easy · theory
Which expression genuinely requires the PRODUCT rule (not just scalar multiplication)?
A. $y = x^2 \sin x$ ✓ Correct
B. $y = \pi e^x$
C. $y = 3x^2$
D. $y = 5\sin x$
Solution: x²·sin x is a product of two functions of x. The others are a constant times a single function, needing only the constant-multiple rule.
Q23 — Product Rule · easy · theory
For y = x·eˣ, choosing u = x and v = eˣ, the product rule gives dy/dx =
A. $x e^x$
B. $e^x$
C. $x e^x - e^x$
D. $x e^x + e^x = e^x(x+1)$ ✓ Correct
Solution: u v' + v u' = x·eˣ + eˣ·1 = eˣ(x + 1).
Q24 — Product Rule · easy · theory
A key advantage of writing a product as u·v before differentiating is that:
A. Each factor can be differentiated separately and combined by the rule ✓ Correct
B. The rule is unnecessary
C. The product becomes a sum automatically
D. The derivative is simply u′v′
Solution: The product rule lets you handle each factor with its own known derivative, then combine — you never just multiply the two derivatives.
Q25 — Product Rule · easy · theory
A common ERROR in applying the product rule is to write (uv)′ as:
A. $uv' + vu'$
B. $vu' + uv'$
C. $u'v'$ (product of the derivatives) ✓ Correct
D. the correct combination
Solution: Writing (uv)′ = u′v′ is wrong; the correct rule has two terms: uv′ + vu′.
Q26 — Product Rule · easy · theory
The product rule can be derived from the definition of the derivative and is valid for:
A. Only trigonometric functions
B. Only polynomial functions
C. Any two differentiable functions ✓ Correct
D. Only when one factor is a constant
Solution: The product rule holds for the product of any two differentiable functions, of any type.
Q27 — Product Rule · easy · theory
For y = (x+1)(x+2), one may either use the product rule OR:
A. It cannot be differentiated another way
B. Use the quotient rule
C. Expand to x² + 3x + 2 and differentiate term by term ✓ Correct
D. Use the chain rule
Solution: Both give 2x + 3. Expanding first is a valid shortcut for simple polynomial products.
Q28 — Product Rule · easy · theory
In the product rule (uv)′ = uv′ + vu′, the two terms are:
A. Subtracted
B. Divided
C. Multiplied together
D. Added, never subtracted ✓ Correct
Solution: The product rule always ADDS the two terms. (Subtraction appears in the quotient rule, a common point of confusion.)
Q29 — Product Rule · easy · theory
To differentiate y = x·ln x, the required derivatives of the factors are:
A. $u' = x$ and $v' = 1$
B. $u' = 0$ and $v' = 1/x$
C. $u' = 1$ (for x) and $v' = 1/x$ (for ln x) ✓ Correct
D. $u' = 1$ and $v' = \ln x$
Solution: For u = x, u′ = 1; for v = ln x, v′ = 1/x. Then y′ = x(1/x) + ln x·1 = 1 + ln x.
Q30 — Product Rule · easy · numerical
If $y = x^2 \sin x$, then $\dfrac{dy}{dx}$ =
A. $2x\cos x$
B. $x^2\cos x - 2x\sin x$
C. $x^2\cos x + 2x\sin x$ ✓ Correct
D. $2x\sin x\cos x$
Solution: u = x², v = sin x: y′ = x²cos x + sin x·2x = x²cos x + 2x sin x.