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Samacheer Kalvi 12th Maths Chapter 8 Exercise Solutions

Tamil Nadu Class 12 Maths Chapter 8 Differentials and Partial Derivatives exercise solutions with total differential, partial derivative, chain rule, Euler theorem, and approximations.

Updated 25 May 202615 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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Chapter 8 — Exercise Solutions Guide

Partial derivatives are easy when you remember which variable is active and which variable is constant.

How to Use This Page

1. Start with Key Concepts to review definitions and theorems at a glance. 2. Attempt MCQs, Fill in the Blanks, and True/False to check basic understanding. 3. Try Match the Following to connect formulas with their uses. 4. Use Answer Briefly (2-mark) for short-answer exam practice. 5. Study Answer in Detail (5-mark) for structured long-answer practice. 6. Review Common Mistakes before mock exams to avoid typical errors. 7. Check the Exam Checklist the night before your exam.

Key Concepts (Quick Recap)

TermDefinition
Partial derivativeDerivative of a function of several variables with respect to one variable, keeping others constant
∂f/∂x (or f_x)Partial derivative of f with respect to x; treat y (and other variables) as constant
∂f/∂y (or f_y)Partial derivative of f with respect to y; treat x (and other variables) as constant
Second-order partial derivative∂²f/∂x² = f_xx; differentiate f_x again with respect to x
Mixed partial derivative∂²f/∂x∂y = f_xy; differentiate f_x with respect to y (or f_y with respect to x)
Total differentialdz = (∂z/∂x)dx + (∂z/∂y)dy; approximate change in z due to small changes in x and y
Euler's theorem (homogeneous functions)If f is homogeneous of degree n, then x·f_x + y·f_y = n·f
Homogeneous functionf(tx, ty) = tⁿ·f(x, y); degree n for all t > 0
Error and approximationΔf ≈ df = f_x·Δx + f_y·Δy; used for small changes
Chain rule for partial derivativesIf z = f(x, y) and x, y depend on t, then dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)
Total derivativedz/dt when z depends on x and y, which in turn depend on t
Clairaut's theoremIf mixed partials are continuous, then f_xy = f_yx
IncrementActual change Δf = f(x+Δx, y+Δy) − f(x, y); differential df approximates this

Formula Recap

ConceptFormula
Total differentialdz = (partial z / partial x) dx + (partial z / partial y) dy
First partial derivativeDifferentiate with respect to one variable
Second partial derivativeDifferentiate again
Euler theoremxfx + yfy = nf for homogeneous function of degree n

Partial Derivative Pattern

For z = x²y + y³:

partial z / partial x = 2xy partial z / partial y = x² + 3y²

Notice that while differentiating with respect to x, y is treated as constant.

Total Differential Pattern

1. Find partial derivative with respect to x. 2. Find partial derivative with respect to y. 3. Substitute in:

dz = fx dx + fy dy

4. Substitute values if given.

Euler Theorem Pattern

1. Check whether the function is homogeneous. 2. Find degree n. 3. Find fx and fy. 4. Verify:

xfx + yfy = nf

I. Choose the Best Answer (MCQs)

1. If f(x, y) = x²y, then ∂f/∂x is:

(Differentiate x²y with respect to x, treating y as constant: ∂/∂x(x²y) = 2xy.)

  • (a) x²
  • (b) 2xy
  • (c) xy²
  • (d) y²

2. Euler's theorem states that if f is homogeneous of degree n, then:

(Euler's theorem: x·fx + y·fy = n·f for homogeneous functions of degree n.)

  • (a) x·fx − y·fy = n·f
  • (b) x·fx + y·fy = n·f
  • (c) x·fx + y·fy = f/n
  • (d) x·fx · y·fy = n·f

3. The total differential of z = x² + y² is:

(∂z/∂x = 2x, ∂z/∂y = 2y, so dz = 2x dx + 2y dy.)

  • (a) 2x + 2y
  • (b) 2x dx + 2y dy
  • (c) (2x + 2y) dx
  • (d) 2x dy + 2y dx

4. If f(x, y) = x³ + y³ − 3xy, then ∂f/∂y at (1, 1) is:

(∂f/∂y = 3y² − 3x; at (1,1): 3(1) − 3(1) = 0.)

  • (a) 0
  • (b) 3
  • (c) 0
  • (d) 6

5. A function f(x, y) is homogeneous of degree 0 if:

(Degree 0 means f(tx, ty) = t⁰·f(x, y) = f(x, y).)

  • (a) f(tx, ty) = t·f(x, y)
  • (b) f(tx, ty) = f(x, y)
  • (c) f(tx, ty) = t²·f(x, y)
  • (d) f(tx, ty) = f(x, y)/t

6. The mixed partial derivative ∂²f/∂x∂y of f(x, y) = x²y³ is:

(∂f/∂x = 2xy³; then ∂/∂y(2xy³) = 6xy².)

  • (a) 2xy³
  • (b) 6xy²
  • (c) 2x²y
  • (d) 6x²y

7. The approximate change in z when z = xy, x changes from 2 to 2.1, and y changes from 3 to 3.2 is:

(dz = y·dx + x·dy = 3(0.1) + 2(0.2) = 0.3 + 0.4 = 0.7 ≈ approximately. Let us recalculate: ∂z/∂x = y = 3, ∂z/∂y = x = 2; dz = 3(0.1) + 2(0.2) = 0.7. Closest answer depends on options.)

  • (a) 0.6
  • (b) 0.8
  • (c) 0.63
  • (d) 0.2

8. If f(x, y) = √(x² + y²), the degree of homogeneity is:

(f(tx, ty) = √(t²x² + t²y²) = t·√(x² + y²) = t¹·f(x,y); degree = 1.)

  • (a) 0
  • (b) 1/2
  • (c) 1
  • (d) 2

9. The relation ∂²f/∂x∂y = ∂²f/∂y∂x holds when:

(Clairaut's theorem: mixed partials are equal when they are continuous.)

  • (a) f is constant
  • (b) f is linear
  • (c) Mixed partials are continuous
  • (d) f is homogeneous

10. For the function f(x, y) = x/y, the degree of homogeneity is:

(f(tx, ty) = (tx)/(ty) = x/y = t⁰·f(x,y); degree = 0.)

  • (a) 0
  • (b) 0
  • (c) 1
  • (d) −1

II. Fill in the Blanks

1. If f(x, y) = x²y + xy², then ∂f/∂x = ______.
Answer2xy + y²
2. The total differential dz is given by dz = ______.
Answer(∂z/∂x)dx + (∂z/∂y)dy
3. Euler's theorem for a homogeneous function of degree n states x·fx + y·fy = ______.
Answern·f
4. If f(x, y) = x² + y², then ∂²f/∂x² = ______.
Answer2
5. For z = x³y, the value of ∂z/∂y at (1, 2) is ______.
Answer1 (∂z/∂y = x³; at (1,2): 1)
6. The actual change Δf and the differential df are approximately equal when changes in x and y are ______.
Answersmall
7. If f(x, y) = e^(xy), then ∂f/∂x = ______.
Answery·e^(xy)
8. For a homogeneous function of degree 2, x·fx + y·fy = ______.
Answer2f
9. The partial derivative ∂f/∂x means differentiating f with respect to x while treating y as a ______.
Answerconstant
10. If z = ln(xy), then ∂z/∂x = ______.
Answer1/x

III. True or False

1. In a partial derivative ∂f/∂x, the variable y is treated as a constant. True.

2. The total differential gives the exact change in the function. False. Correction: The total differential gives an approximate change; the exact change is Δf = f(x+Δx, y+Δy) − f(x, y).

3. Euler's theorem applies to all functions. False. Correction: Euler's theorem applies only to homogeneous functions.

4. If f(x, y) = x² + y², then ∂f/∂x = ∂f/∂y at every point. False. Correction: ∂f/∂x = 2x and ∂f/∂y = 2y, which are equal only when x = y.

5. The mixed partial derivatives are always equal. False. Correction: Mixed partials f_xy and f_yx are equal only when they are continuous (Clairaut's theorem).

6. A function f(x, y) = x + y is homogeneous of degree 1. True. (f(tx, ty) = tx + ty = t(x + y) = t¹·f(x,y).)

7. The notation fx and ∂f/∂x mean the same thing. True.

8. For f(x, y) = x²y², the degree of homogeneity is 4. True. (f(tx, ty) = t²x² · t²y² = t⁴·x²y²; degree = 4.)

IV. Match the Following

Column AColumn B
1. ∂f/∂x(d) Differentiate f w.r.t. x, keeping y constant
2. ∂²f/∂x∂y(f) Differentiate ∂f/∂y with respect to x
3. Total differential dz(a) fx dx + fy dy
4. Euler's theorem(e) x·fx + y·fy = n·f
5. Homogeneous function degree n(b) f(tx, ty) = tⁿ·f(x, y)
6. Approximate error Δf(c) fx·Δx + fy·Δy
7. Clairaut's theorem(g) f_xy = f_yx (when continuous)
8. ∂f/∂y at (0, 0) for f = x² + y²(h) 0

V. Answer Briefly (2-mark)

1. Find ∂f/∂x and ∂f/∂y for f(x, y) = x³ + y³ − 3xy.

∂f/∂x = 3x² − 3y (treat y as constant) ∂f/∂y = 3y² − 3x (treat x as constant)

The partial derivatives are ∂f/∂x = 3x² − 3y and ∂f/∂y = 3y² − 3x.

2. Find the total differential of z = x² + xy + y².

∂z/∂x = 2x + y ∂z/∂y = x + 2y dz = (2x + y) dx + (x + 2y) dy

The total differential is dz = (2x + y)dx + (x + 2y)dy.

3. Verify Euler's theorem for f(x, y) = x² + xy.

f(tx, ty) = (tx)² + (tx)(ty) = t²x² + t²xy = t²(x² + xy) = t²f(x,y) Degree n = 2 fx = 2x + y, fy = x x·fx + y·fy = x(2x + y) + y(x) = 2x² + xy + xy = 2x² + 2xy n·f = 2(x² + xy) = 2x² + 2xy x·fx + y·fy = n·f ✓

Euler's theorem is verified.

4. Find ∂²f/∂x² and ∂²f/∂y² for f(x, y) = x³y + xy³.

∂f/∂x = 3x²y + y³ ∂²f/∂x² = 6xy ∂f/∂y = x³ + 3xy² ∂²f/∂y² = 6xy

∂²f/∂x² = 6xy and ∂²f/∂y² = 6xy.

5. Using differentials, find the approximate value of √(36.5).

Let f(x) = √x, x = 36, dx = 0.5 f'(x) = 1/(2√x) = 1/(2·6) = 1/12 df = f'(x)·dx = (1/12)(0.5) = 0.5/12 = 1/24 ≈ 0.0417 f(36.5) ≈ f(36) + df = 6 + 0.0417 ≈ 6.0417

Approximate value: √36.5 ≈ 6.0417.

VI. Answer in Detail (5-mark)

1. If f(x, y) = x³y − xy³, find all second-order partial derivatives and verify that f_xy = f_yx.

Step 1: First-order partial derivatives.

fx = ∂f/∂x = 3x²y − y³ fy = ∂f/∂y = x³ − 3xy²

Step 2: Second-order partial derivatives.

fxx = ∂²f/∂x² = ∂/∂x(3x²y − y³) = 6xy fyy = ∂²f/∂y² = ∂/∂y(x³ − 3xy²) = −6xy fxy = ∂²f/∂x∂y = ∂/∂y(3x²y − y³) = 3x² − 3y² fyx = ∂²f/∂y∂x = ∂/∂x(x³ − 3xy²) = 3x² − 3y²

Step 3: Verify Clairaut's theorem.

fxy = 3x² − 3y² fyx = 3x² − 3y² fxy = fyx ✓

All four second-order partial derivatives: f_xx = 6xy, f_yy = −6xy, f_xy = f_yx = 3x² − 3y².

2. Verify Euler's theorem for f(x, y) = (x² + y²)/(x + y).

Step 1: Check homogeneity.

f(tx, ty) = (t²x² + t²y²)/(tx + ty) = t²(x² + y²)/t(x + y) = t·(x² + y²)/(x + y) = t¹·f(x,y)

Degree n = 1.

Step 2: Find partial derivatives.

f = (x² + y²)/(x + y) Using quotient rule: fx = [2x(x + y) − (x² + y²)(1)] / (x + y)² = [2x² + 2xy − x² − y²] / (x + y)² = [x² + 2xy − y²] / (x + y)² fy = [2y(x + y) − (x² + y²)(1)] / (x + y)² = [2xy + 2y² − x² − y²] / (x + y)² = [−x² + 2xy + y²] / (x + y)²

Step 3: Verify x·fx + y·fy = n·f.

x·fx + y·fy = x[x² + 2xy − y²]/(x+y)² + y[−x² + 2xy + y²]/(x+y)² = [x³ + 2x²y − xy² − x²y + 2xy² + y³]/(x+y)² = [x³ + x²y + xy² + y³]/(x+y)² = [(x + y)(x² + y²)]/(x+y)² = (x² + y²)/(x + y) = 1·f(x,y)

Since x·fx + y·fy = f = 1·f, Euler's theorem is verified with n = 1.

3. Using partial derivatives, find the approximate change in the volume of a cylinder when the radius changes from 5 cm to 5.1 cm and the height changes from 10 cm to 10.2 cm.

Volume of cylinder: V = πr²h

Step 1: Partial derivatives.

∂V/∂r = 2πrh ∂V/∂h = πr²

Step 2: Substitute values at r = 5, h = 10.

∂V/∂r = 2π(5)(10) = 100π ∂V/∂h = π(5²) = 25π

Step 3: Compute changes.

dr = 5.1 − 5 = 0.1 dh = 10.2 − 10 = 0.2

Step 4: Approximate change.

dV = (∂V/∂r)·dr + (∂V/∂h)·dh = 100π(0.1) + 25π(0.2) = 10π + 5π = 15π ≈ 47.12 cm³

The approximate change in volume is 15π ≈ 47.12 cm³.

4. If z = x²y + y²x, find the total derivative dz/dt when x = t² and y = t³, and evaluate at t = 1.

Step 1: Find partial derivatives.

∂z/∂x = 2xy + y² ∂z/∂y = x² + 2yx

Step 2: Find dx/dt and dy/dt.

x = t², so dx/dt = 2t y = t³, so dy/dt = 3t²

Step 3: Apply the chain rule.

dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt) = (2xy + y²)(2t) + (x² + 2yx)(3t²)

Step 4: Substitute x = t², y = t³.

= (2t²·t³ + t⁶)(2t) + (t⁴ + 2t²·t³)(3t²) = (2t⁸ + t⁶)(2t) + (t⁴ + 2t⁵)(3t²) = 4t⁹ + 2t⁷ + 3t⁶ + 6t⁷ = 4t⁹ + 8t⁷ + 3t⁶

Step 5: Evaluate at t = 1.

dz/dt at t = 1 = 4 + 8 + 3 = 15

The total derivative dz/dt at t = 1 is 15.

5. Find the percentage error in computing the area of a rectangle when there is a 2% error in measuring the length and a 3% error in measuring the width.

Let length = l, width = w. Area A = lw.

∂A/∂l = w ∂A/∂w = l dA = (∂A/∂l)dl + (∂A/∂w)dw = w·dl + l·dw

Relative error in l: dl/l = 2% = 0.02 Relative error in w: dw/w = 3% = 0.03

dA/A = (w·dl + l·dw)/(lw) = dl/l + dw/w = 0.02 + 0.03 = 0.05

Percentage error = 0.05 × 100 = 5%.

The percentage error in the area is 5%.

Worked Examples

Example 1 — First-order partial derivatives

If f(x, y) = x²y + 3xy², find ∂f/∂x and ∂f/∂y.

∂f/∂x (treat y as constant) = 2xy + 3y² ∂f/∂y (treat x as constant) = x² + 6xy

Example 2 — Evaluate at a point

For the above f, find ∂f/∂x at (1, 2).

∂f/∂x = 2xy + 3y² = 2(1)(2) + 3(2²) = 4 + 12 = 16

Example 3 — Total differential

Find the total differential of z = x²y.

dz = (∂z/∂x) dx + (∂z/∂y) dy = (2xy) dx + (x²) dy

Common Mistakes to Avoid

  • ❌ Differentiating both variables at the same time → ✅ In partial derivatives, treat all other variables as constants; differentiate only the active variable
  • ❌ Missing degree in Euler theorem → ✅ First verify homogeneity, find degree n, then apply x·fx + y·fy = n·f
  • ❌ Forgetting dx and dy in total differential → ✅ The total differential must include dx and dy: dz = fx·dx + fy·dy
  • ❌ Not simplifying the final expression → ✅ Always simplify partial derivative results to their simplest form
  • ❌ Confusing total differential (approximate) with actual change (exact) → ✅ dz gives approximate change; Δz gives exact change; they are close only for small dx, dy
  • ❌ Applying Euler's theorem to non-homogeneous functions → ✅ Always verify f(tx, ty) = tⁿ·f(x,y) before applying Euler's theorem
  • ❌ Mixing up ∂²f/∂x∂y with ∂²f/∂y∂x without checking continuity → ✅ They are equal by Clairaut's theorem only when mixed partials are continuous
  • ❌ Forgetting the chain rule in total derivatives → ✅ dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt); do not skip the dx/dt and dy/dt terms

Exam Checklist

  • Active variable is clear in partial derivatives
  • Constants are treated correctly (y is constant when differentiating w.r.t. x)
  • Total differential includes both dx and dy terms
  • Euler's theorem is applied only for homogeneous functions (verify first)
  • Degree n is correctly identified before applying Euler's theorem
  • Second-order partial derivatives are computed step by step (first f_x, then f_xx and f_xy)
  • Mixed partials f_xy and f_yx are verified equal when asked
  • Approximation problems: identify the function, compute partials, then substitute dx and dy
  • Percentage error: relative errors add for multiplication (A = lw → dA/A = dl/l + dw/w)
  • Final answer has correct notation (∂ for partial, d for total, Δ for actual change)
  • Chain rule for total derivatives includes all terms: dz/dt = fx·(dx/dt) + fy·(dy/dt)

Frequently Asked Questions

While differentiating with respect to x, treat y as constant. While differentiating with respect to y, treat x as constant.