68 free MCQs on Limits and Derivatives with worked answers and explanations. Rate of change, differentiation rules, and applications
Take the timed Limits and Derivatives chapterwise test →Below are 68 practice questions on Limits and Derivatives, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Limits and Derivatives notes.
As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.
Q1.
d/dx (x²) =
Answer: C. 2x
Why: Power rule: d/dx (xⁿ) = nxⁿ⁻¹. So d/dx(x²) = 2x.
Q2.
d/dx (sin x) =
Answer: A. cos x
Why: The derivative of sin x is cos x.
Q3.
d/dx (cos x) =
Answer: B. -sin x
Why: The derivative of cos x is -sin x (note the negative sign).
Q4.
d/dx (constant) =
Answer: B. 0
Why: The derivative of any constant is zero. Constants do not change.
Q5.
d/dx (eˣ) =
Answer: C. eˣ
Why: The exponential function eˣ is its own derivative: d/dx(eˣ) = eˣ.
Q6.
The derivative of f(x) represents:
Answer: C. Instantaneous rate of change
Why: The derivative gives the instantaneous rate of change of the function at any point.
Q7.
d/dx (x³ + 2x + 1) =
Answer: A. 3x² + 2
Why: Differentiate each term: d/dx(x³) = 3x², d/dx(2x) = 2, d/dx(1) = 0. Total: 3x² + 2.
Q8.
lim (x→0) of sin x / x =
Answer: C. 1
Why: Standard limit: lim(x→0) sin x / x = 1. This is a fundamental calculus result.
Q9.
d/dx (ln x) =
Answer: B. 1/x
Why: d/dx(ln x) = 1/x for x > 0.
Q10.
d/dx (tan x) =
Answer: C. sec²x
Why: d/dx(tan x) = sec²x.
Q11.
If f(x) = 5x, then f'(x) =
Answer: C. 5
Why: d/dx(5x) = 5 (constant coefficient rule).
Q12.
d/dx (xⁿ) =
Answer: C. nxⁿ⁻¹
Why: Power rule: d/dx(xⁿ) = n × xⁿ⁻¹.
Q13.
At a maximum, the derivative of a function is:
Answer: C. Zero (first derivative test)
Why: At a maximum or minimum, f'(x) = 0 (critical point). Check sign of f'' to distinguish.
Q14.
lim (x→2) of x² = ?
Answer: C. 4
Why: Since x² is continuous, lim(x→2) x² = 2² = 4.
Q15.
The derivative of x with respect to x is:
Answer: C. 1
Why: d/dx(x) = d/dx(x¹) = 1×x⁰ = 1.
Q16.
d/dx (3x² + 5x) =
Answer: A. 6x + 5
Why: d/dx(3x²) = 6x, d/dx(5x) = 5. Total: 6x + 5.
Q17.
If y = x², then dy/dx at x = 3 is:
Answer: B. 6
Why: dy/dx = 2x. At x = 3: dy/dx = 2(3) = 6.
Q18.
The slope of the tangent to y = x² at point (2,4) is:
Answer: B. 4
Why: Slope = dy/dx = 2x. At x = 2: slope = 4.
Q19.
d/dx (1/x) =
Answer: B. -1/x²
Why: 1/x = x⁻¹. d/dx(x⁻¹) = -1 × x⁻² = -1/x².
Q20.
lim (x→3) of (x - 3)/(x² - 9) =
Answer: B. 1/6
Why: Factor: x²-9 = (x-3)(x+3). So (x-3)/[(x-3)(x+3)] = 1/(x+3). As x→3: 1/6.
Q21.
Find d/dx (x² sin x).
Answer: A. 2x sin x + x² cos x
Why: Product rule: d/dx(fg) = f'g + fg'. f=x², f'=2x; g=sinx, g'=cosx. Answer: 2x sinx + x² cosx.
Q22.
Find d/dx (sin x / x).
Answer: A. (x cos x - sin x)/x²
Why: Quotient rule: (f'g - fg')/g². f=sinx, f'=cosx; g=x, g'=1. Result: (x cosx - sinx)/x².
Q23.
lim(x→0) (1-cos x)/x² =
Answer: C. 1/2
Why: Using L'Hopital or expansion: (1-cosx) = x²/2 - x⁴/24+... So (1-cosx)/x² = 1/2 as x→0.
Q24.
Find d/dx [f(g(x))] when f(u) = u² and g(x) = sin x.
Answer: A. 2sin x cos x
Why: Chain rule: f'(g(x))×g'(x) = 2sin(x) × cos(x) = 2sinx cosx = sin 2x.
Q25.
If y = (2x+3)⁵, find dy/dx.
Answer: B. 10(2x+3)⁴
Why: Chain rule: 5(2x+3)⁴ × d/dx(2x+3) = 5(2x+3)⁴ × 2 = 10(2x+3)⁴.
Q26.
Find the critical points of f(x) = x³ - 3x² + 2.
Answer: A. x = 0, 2
Why: f'(x) = 3x² - 6x = 3x(x-2) = 0. Critical points: x = 0 and x = 2.
Q27.
Second derivative (f'') > 0 at a critical point means:
Answer: B. Minimum
Why: f''(x) > 0 at critical point: concave up, so it is a local minimum.
Q28.
lim(x→∞) of (3x² + 5)/(x² + 1) =
Answer: B. 3
Why: Divide by x²: (3 + 5/x²)/(1 + 1/x²). As x→∞: 3/1 = 3.
Q29.
The equation of tangent to y = x² at (1,1) is:
Answer: A. y = 2x - 1
Why: dy/dx = 2x. At (1,1): slope = 2. Tangent: y-1 = 2(x-1), y = 2x-1.
Q30.
d/dx (log₁₀ x) =
Answer: D. 1/(x ln 10)
Why: d/dx(log₁₀ x) = 1/(x ln 10) = (log₁₀ e)/x. Both forms are equivalent.
Q31.
If f(x) = x³ - 6x² + 9x, the function is increasing when:
Answer: A. x < 1 or x > 3
Why: f'(x) = 3x² - 12x + 9 = 3(x-1)(x-3) > 0 when x < 1 or x > 3.
Q32.
The derivative of sec x is:
Answer: A. sec x tan x
Why: d/dx(sec x) = sec x tan x.
Q33.
d/dx (x<sup>x</sup>) =
Answer: C. x<sup>x</sup> (1 + ln x)
Why: Take ln: y = x ln x. Differentiate: y'/y = ln x + 1. y' = y(1+ln x) = x<sup>x</sup>(1+ln x).
Q34.
The derivative of sin⁻¹(x) is:
Answer: A. 1/√(1-x²)
Why: d/dx(sin⁻¹ x) = 1/√(1-x²).
Q35.
lim(x→0) (eˣ - 1)/x =
Answer: B. 1
Why: Standard limit: lim(x→0) (eˣ-1)/x = 1.
Q36.
If y = ln(sin x), find dy/dx.
Answer: A. cot x
Why: dy/dx = 1/sin(x) × cos(x) = cos x / sin x = cot x.
Q37.
A particle moves as x = t³ - 3t. Velocity at t = 2 is:
Answer: B. 9
Why: v = dx/dt = 3t² - 3. At t=2: v = 12 - 3 = 9.
Q38.
Rolle's theorem requires f to be:
Answer: B. Continuous on [a,b], differentiable on (a,b), and f(a) = f(b)
Why: Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists c in (a,b) with f'(c) = 0.
Q39.
d/dx (tan⁻¹ x) =
Answer: A. 1/(1+x²)
Why: d/dx(tan⁻¹ x) = 1/(1+x²).
Q40.
If f(x) = x² - 4x + 5, find the minimum value.
Answer: A. 1
Why: f'(x) = 2x - 4 = 0, x = 2. f(2) = 4 - 8 + 5 = 1. Minimum value = 1.
Q41.
Evaluate: lim(x→0) [sin(x)/x]<sup>1/x²</sup> using limits.
Answer: A. e<sup>-1/6</sup>
Why: 1^∞ indeterminate form: take ln L = lim[ln(sinx/x)]/x². Using Taylor: sinx/x = 1 − x²/6 + …, so ln(sinx/x) ≈ −x²/6. Hence ln L = −1/6, giving L = e<sup>−1/6</sup>.
Q42.
lim(x→0) (1/x² - cosec²x) =
Answer: A. -1/3
Why: Write as (sin²x − x²)/(x² sin²x). Using sinx ≈ x − x³/6: sin²x ≈ x² − x⁴/3, so sin²x − x² ≈ −x⁴/3. Denominator ≈ x²·x² = x⁴. Limit = (−x⁴/3)/x⁴ = −1/3. The negative sign is essential - a very common sign error on this problem.
Q43.
If y = (sin x)<sup>tan x</sup>, find dy/dx.
Answer: A. (sinx)<sup>tanx</sup> × [sec²x × ln(sinx) + 1]
Why: Take ln: y = tanx × ln(sinx). Differentiate: y'/y = sec²x × ln(sinx) + tanx × cosx/sinx = sec²x lnsinx + 1. y' = y × [sec²x lnsinx + 1].
Q44.
The Mean Value Theorem states that for f on [a,b]:
Answer: C. f(b)-f(a) = f'(c)(b-a) for some c in (a,b)
Why: MVT (Lagrange): if f is continuous on [a,b] and differentiable on (a,b), there exists c in (a,b) such that f'(c) = [f(b)-f(a)]/(b-a).
Q45.
Using L'Hopital: lim(x→0) (eˣ - e⁻ˣ - 2x)/(x - sinx) =
Answer: C. 2
Why: Applying L'Hopital twice: Num' = eˣ+e⁻ˣ-2; Den' = 1-cosx. Both 0 at x=0. Again: Num'' = eˣ-e⁻ˣ; Den'' = sinx. Still 0. Again: eˣ+e⁻ˣ / cosx → 2/1 = 2.
Q46.
The inflection point of f(x) = x⁴ - 4x³ is at:
Answer: C. x = 0 and x = 2
Why: Compute f''(x): f'(x) = 4x³ − 12x², f''(x) = 12x² − 24x = 12x(x−2). f'' = 0 at x = 0 and x = 2. Sign of f'' changes at both points (−/+/−), confirming inflection points at x = 0 and x = 2.
Q47.
Differentiate: y = cos⁻¹(2x²-1) with respect to x.
Answer: B. -2/√(1-x²)
Why: Substitution: let x = cosθ, so 2x²−1 = 2cos²θ−1 = cos2θ. Then y = cos⁻¹(cos2θ) = 2θ = 2cos⁻¹x. Differentiating: dy/dx = 2·(−1/√(1−x²)) = −2/√(1−x²).
Q48.
lim(x→1) (x<sup>n</sup> - 1)/(x - 1) =
Answer: B. n
Why: This is the definition of derivative of x<sup>n</sup> at x=1. Or factor: (x<sup>n</sup>-1)/(x-1) = x<sup>n-1</sup>+x<sup>n-2</sup>+...+1 (n terms). At x=1: = n.
Q49.
If f(x) = x|x|, then f'(0) is:
Answer: A. 0
Why: f(x) = x² (x>0), −x² (x<0). Using first-principles: RHD = lim(h→0⁺) h²/h = 0; LHD = lim(h→0⁻) (−h²)/h = 0. Both one-sided derivatives equal 0, so f'(0) = 0.
Q50.
Taylor series expansion of eˣ around x=0 gives the n-th derivative value at 0 as:
Answer: B. 1
Why: Taylor series: eˣ = Σ(xⁿ/n!). The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, and here it equals 1/n!. So f⁽ⁿ⁾(0) = 1 for all n ≥ 0, confirmed since d<sup>n</sup>/dx<sup>n</sup>(eˣ) = eˣ and e⁰ = 1.
Q51.
Find the equation of normal to curve xy = 4 at point (2,2).
Answer: A. y = x
Why: Implicit differentiation of xy = 4: y + xy' = 0, so y' = −y/x = −1 at (2,2). Tangent slope = −1, so normal slope = 1. Normal line: y − 2 = 1·(x − 2), giving y = x.
Q52.
Newton-Leibniz rule for d/dx [integral from a to g(x) of f(t) dt] =
Answer: C. f(g(x)) × g'(x)
Why: By the Fundamental Theorem of Calculus (Part 1), d/dx[∫_a<sup>u</sup> f(t)dt] = f(u). Applying the chain rule with u = g(x): d/dx[∫_a^{g(x)} f(t)dt] = f(g(x))·g'(x).
Q53.
If f''(a) > 0 and f'(a) = 0 and f''(a) exists, then f(a) is:
Answer: B. Local minimum
Why: Second derivative test: f'(a) = 0 identifies a critical point. f''(a) > 0 means f' is increasing through zero, so f changes from decreasing to increasing at a. Therefore f(a) is a local minimum.
Q54.
Differentiate: y = sin(cos(tan x))
Answer: A. -cos(cos(tanx)) sin(tanx) sec²x
Why: Chain rule (3 layers): dy/dx = cos(cos(tanx)) · d/dx[cos(tanx)]. Next layer: d/dx[cos(tanx)] = −sin(tanx)·sec²x. Combining: dy/dx = −cos(cos(tanx))·sin(tanx)·sec²x.
Q55.
If x = a cos t, y = a sin t (parametric), dy/dx =
Answer: A. -cot t
Why: Parametric differentiation: dx/dt = −a sin t, dy/dt = a cos t. dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(−a sin t) = −cos t/sin t = −cot t. Note: the correct result is −cot t, not −tan t.
Q56.
Cauchy Mean Value Theorem applies to functions f and g. It states: there exists c in (a,b) such that:
Answer: A. f'(c)/g'(c) = [f(b)-f(a)]/[g(b)-g(a)]
Why: Cauchy MVT generalises Lagrange's MVT to two functions: for continuous f, g on [a,b], differentiable on (a,b) with g'(x) ≠ 0, there exists c ∈ (a,b) such that f'(c)/g'(c) = [f(b)−f(a)]/[g(b)−g(a)].
Q57.
The value of lim (x→0) (sin x)/x is:
Answer: A. 1
Why: This is a standard limit: lim (x→0) (sin x)/x = 1.
Q58.
The derivative of x³ with respect to x is:
Answer: A. 3x²
Why: By the power rule, d(xⁿ)/dx = n·xⁿ⁻¹, so d(x³)/dx = 3x².
Q59.
The value of lim (x→2) (x² − 4)/(x − 2) is:
Answer: A. 4
Why: Factor: (x−2)(x+2)/(x−2) = x + 2, so the limit as x→2 is 4.
Q60.
The derivative of sin x with respect to x is:
Answer: A. cos x
Why: d(sin x)/dx = cos x.
Q61.
The limit as n → ∞ of (1/n)·[(n+1)(n+2)...(2n)]<sup>1/n</sup> is:
Answer: B. 4/e
Why: Taking logs converts this to ∫₀¹ ln(1+x) dx = 2ln2 − 1, so the limit is e<sup>2ln2 − 1</sup> = 4/e.
Q62.
The limit as x → 0 of (cos x)<sup>1/x²</sup> is:
Answer: B. e<sup>−1/2</sup>
Why: ln(cosx)/x² → −1/2, so the limit is e<sup>−1/2</sup>.
Q63.
The limit as x → 0 of sin5x / sin3x is:
Answer: C. 5/3
Why: Both sines behave linearly near 0, giving 5x/3x = 5/3.
Q64.
The limit as x → 0 of (e<sup>2x</sup> − 1) / ln(1 + 3x) is:
Answer: A. 2/3
Why: Numerator ~ 2x and denominator ~ 3x, so the ratio tends to 2/3.
Q65.
The limit as x → 2 of (x³ − 8)/(x² − 4) is:
Answer: B. 3
Why: Factor: (x−2)(x²+2x+4)/[(x−2)(x+2)] → (4+4+4)/4 = 3.
Q66.
The limit as x → 0 of (1 + 2x)<sup>1/x</sup> is:
Answer: B. e²
Why: (1 + 2x)<sup>1/x</sup> = [(1 + 2x)<sup>1/2x</sup>]² → e².
Q67.
The derivative of x<sup>x</sup> at x = 1 is:
Answer: B. 1
Why: d/dx x<sup>x</sup> = x<sup>x</sup>(1 + lnx); at x = 1 this is 1·(1 + 0) = 1.
Q68.
The limit as x → 0 of (√(1 + x) − √(1 − x)) / x is:
Answer: C. 1
Why: Numerator ~ (1 + x/2) − (1 − x/2) = x, so the ratio tends to 1.