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CBSE · Class 12 · Mathematics

Implicit and Inverse Trigonometric Differentiation (Ch 5: 5.3.2–5.3.3)

2 concepts · 16 practice questions with answers · Free to read

Implicit differentiation

From daily life

On a circle, y is not given as a neat formula in x. Yet at every point the circle has a slope. We find it by differentiating the whole equation.

Simple definition

When y is tied to x by an equation that is not solved for y, differentiate both sides with respect to x, remembering that y depends on x: d/dx (y²) = 2y (dy/dx). Then solve for dy/dx.

Formula: d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx), d/dx (xy) = y + x (dy/dx)

Rule 1: Differentiate and solve — Collect the dy/dx terms on one side.

  • Easy: x² + y² = 25 gives 2x + 2y y′ = 0, so y′ = −x / y.
  • Medium: xy = 6 gives y + x y′ = 0, so y′ = −y / x.
  • Exam-type: x³ + y³ = 3xy gives y′ = (y − x²) / (y² − x).

Rule 2: Slope at a point — Substitute x and y after finding dy/dx.

  • Easy: x² + y² = 25 at (3, 4): slope = −3/4.
  • Medium: x² + xy + y² = 7 at (1, 2): y′ = −(2x + y) / (x + 2y) = −4/5.
  • Exam-type: y² = 4x at (1, 2): 2y y′ = 4, so y′ = 2 / y = 1.
Remember

Every y term needs a factor of dy/dx when differentiated.

Common mistake

Differentiating y² as 2y and forgetting dy/dx.

Memory clue

y is a function of x in disguise: always attach y′.

Practice questions

Easy
  1. 1Find dy/dx if x² + y² = 16.
  2. 2Find dy/dx if xy = 10.
  3. 3Find dy/dx if x² + 2y² = 9.
Medium
  1. 4Find dy/dx if x³ + y³ = 8.
  2. 5Find dy/dx if y² = 8x.
  3. 6Find dy/dx if sin y = x.
Challenging
  1. 7Find the slope of x² + xy + y² = 7 at the point (2, 1).
  2. 8Find dy/dx if x² + y² = sin y.
Show answers (try them yourself first)
  1. 1−x / y
  2. 2−y / x
  3. 3−x / (2y)
  4. 4−x² / y²
  5. 54 / y
  6. 6cos y × y′ = 1, so y′ = 1 / cos y
  7. 7y′ = −(2x + y) / (x + 2y) = −5/4
  8. 82x + 2y y′ = cos y y′, so y′ = 2x / (cos y − 2y)

Derivatives of inverse trigonometric functions

From daily life

The inverse of sine returns an angle from a ratio. How fast the angle changes when the ratio changes is the derivative.

Simple definition

For |x| < 1, d/dx sin⁻¹x = 1/√(1 − x²) and d/dx cos⁻¹x = −1/√(1 − x²). For all x, d/dx tan⁻¹x = 1/(1 + x²) and d/dx cot⁻¹x = −1/(1 + x²). For |x| > 1, d/dx sec⁻¹x = 1/(|x|√(x² − 1)) and d/dx cosec⁻¹x = −1/(|x|√(x² − 1)).

Formula: d/dx tan⁻¹x = 1/(1 + x²), d/dx sin⁻¹x = 1/√(1 − x²)

Rule 1: Use the formula with the chain rule — Replace x by the inside function and multiply by its derivative.

  • Easy: d/dx tan⁻¹(2x) = 2 / (1 + 4x²).
  • Medium: d/dx sin⁻¹(x²) = 2x / √(1 − x⁴).
  • Exam-type: d/dx cos⁻¹(3x) = −3 / √(1 − 9x²).

Rule 2: Simplify first — Use identities to make the inside simple.

  • Easy: sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹x for |x| < 1/√2, so the derivative is 2/√(1 − x²).
  • Medium: sin⁻¹(3x − 4x³) = 3 sin⁻¹x for |x| ≤ 1/2, so the derivative is 3/√(1 − x²).
  • Exam-type: cos⁻¹((1 − x²)/(1 + x²)) = 2 tan⁻¹x for x ≥ 0, so the derivative is 2/(1 + x²).
Remember

Sin⁻¹ and tan⁻¹ have positive derivatives; the "co" partners have the same expression with a minus sign.

Common mistake

Writing d/dx sin⁻¹x = 1/sin x.

Memory clue

Co-functions carry the minus sign.

Practice questions

Easy
  1. 1Differentiate sin⁻¹x.
  2. 2Differentiate tan⁻¹(5x).
  3. 3Differentiate cos⁻¹x.
Medium
  1. 4Differentiate sin⁻¹(3x).
  2. 5Differentiate cot⁻¹(x²).
  3. 6Differentiate tan⁻¹(√x).
Challenging
  1. 7Find d/dx (sin⁻¹x + cos⁻¹x).
  2. 8Differentiate sin⁻¹(2x√(1 − x²)) after simplifying.
Show answers (try them yourself first)
  1. 11 / √(1 − x²)
  2. 25 / (1 + 25x²)
  3. 3−1 / √(1 − x²)
  4. 43 / √(1 − 9x²)
  5. 5−2x / (1 + x⁴)
  6. 6[1 / (1 + x)] × [1 / (2√x)] = 1 / (2√x (1 + x))
  7. 70, since the sum is the constant π/2
  8. 82 / √(1 − x²) for −1/√2 < x < 1/√2