Exercise 5.2 answers: Continuity and Differentiability

Open NCERT p. 122Matches NCERT’s answer

1. Treat this as an outer rule (sine) applied to an inner rule (x² + 5).
2. The derivative of sin t is cos t, so at the inner value it gives cos(x² + 5).
3. The derivative of the inner rule x² + 5 is 2x.
4. By the chain rule, multiply the two: dy/dx = 2x cos(x² + 5).

Answerdy/dx = 2x cos(x² + 5).

Watch this explained “The drill”, 8:39 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. Outer rule: cosine. Inner rule: sin x.
2. The derivative of cos t is −sin t, so at the inner value it gives −sin(sin x).
3. The derivative of the inner rule sin x is cos x.
4. By the chain rule, dy/dx = −sin(sin x) · cos x.

Answerdy/dx = −cos x · sin(sin x).

Watch this explained “The drill”, 8:39 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. Outer rule: sine. Inner rule: ax + b.
2. The derivative of sin t is cos t, so at the inner value it gives cos(ax + b).
3. The derivative of the inner rule ax + b, with respect to x, is a.
4. By the chain rule, dy/dx = a cos(ax + b).

Answerdy/dx = a cos(ax + b).

Watch this explained “The drill”, 8:39 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. This is three rules stacked: innermost √x, middle tan, outer sec.
2. The derivative of √x is 1/(2√x).
3. The derivative of tan t is sec² t, so at √x it gives sec²(√x).
4. The derivative of sec t is sec t · tan t, so at tan(√x) it gives sec(tan √x) · tan(tan √x).
5. Multiply the three factors together: dy/dx = sec(tan √x) · tan(tan √x) · sec²(√x) · 1/(2√x).

Answerdy/dx = [sec(tan √x) · tan(tan √x) · sec²(√x)] / (2√x).

Watch this explained “Three stages, three factors”, 6:55 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. Write y as u/v with u = sin(ax + b) and v = cos(cx + d).
2. By the chain rule, u′ = a cos(ax + b) and v′ = −c sin(cx + d).
3. By the quotient rule, dy/dx = (u′v − uv′) / v².
4. Substitute: dy/dx = [a cos(ax+b) cos(cx+d) − sin(ax+b) · (−c sin(cx+d))] / cos²(cx+d).
5. Simplify the sign: dy/dx = [a cos(ax+b) cos(cx+d) + c sin(ax+b) sin(cx+d)] / cos²(cx+d).

Answerdy/dx = [a cos(ax+b) cos(cx+d) + c sin(ax+b) sin(cx+d)] / cos²(cx+d).

Watch this explained “The drill”, 8:39 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. Write y as a product u · v, with u = cos(x³) and v = sin²(x⁵).
2. For u′: outer cosine, inner x³; the derivative of x³ is 3x², so u′ = −3x² sin(x³).
3. For v′: v = [sin(x⁵)]², so by the chain rule v′ = 2 sin(x⁵) cos(x⁵) · (derivative of x⁵) = 10x⁴ sin(x⁵) cos(x⁵).
4. By the product rule, dy/dx = u′v + uv′.
5. Substitute: dy/dx = −3x² sin(x³) sin²(x⁵) + 10x⁴ sin(x⁵) cos(x⁵) cos(x³).

Answerdy/dx = −3x² sin(x³) sin²(x⁵) + 10x⁴ sin(x⁵) cos(x⁵) cos(x³).

Watch this explained “The drill”, 8:39 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. Write y = 2√(cot(x²)). It has three layers: x² sits inside cot, cot(x²) sits inside the square root, and the result is doubled.
2. Outer layer: the derivative of 2√u is 2 × 1/(2√u) = 1/√u. With u = cot(x²), this factor is 1/√(cot(x²)).
3. Middle layer: the derivative of cot v is −cosec²v. With v = x², this factor is −cosec²(x²).
4. Inner layer: the derivative of x² is 2x.
5. Multiply the three factors: dy/dx = 1/√(cot(x²)) × (−cosec²(x²)) × 2x = −2x·cosec²(x²)/√(cot(x²)).
6. Since cosec²(x²) = 1/sin²(x²), this is −2x / [sin²(x²)·√(cot(x²))].

Answerdy/dx = −2x / [sin²(x²) · √(cot(x²))]

Watch this explained “Three stages, three factors”, 6:55 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122Matches NCERT’s answer

1. This is cos of an inner function √x — two layers.
2. Outer layer: the derivative of cos(u) is −sin(u).
3. Inner layer: the derivative of √x is 1/(2√x).
4. Multiply the two: −sin(√x) × 1/(2√x).

Answerdy/dx = −sin(√x) / (2√x)

Watch this explained “Name the inner rule first”, 7:45 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 122One way to think about it

1. For f to be differentiable at x = 1, its left-hand and right-hand derivatives there must be equal.
2. Left-hand derivative: for small negative h, |1 + h − 1| = |h| = −h, so [f(1+h) − f(1)]/h = −h/h = −1.
3. Right-hand derivative: for small positive h, |1 + h − 1| = |h| = h, so [f(1+h) − f(1)]/h = h/h = 1.
4. The left-hand derivative (−1) is not equal to the right-hand derivative (1).

In shortSince the left-hand derivative (−1) and the right-hand derivative (1) at x = 1 are different, f is not differentiable at x = 1.

Watch this explained “The modulus at nought”, 8:49 into Differentiability, and why it forces continuity while the reverse fails

Open NCERT p. 122One way to think about it

1. A function must be continuous at a point in order to be differentiable there.
2. At x = 1: as x approaches 1 from the left, [x] = 0, but f(1) = [1] = 1. The left-hand limit does not equal f(1), so f is discontinuous at x = 1.
3. At x = 2: as x approaches 2 from the left, [x] = 1, but f(2) = [2] = 2. The left-hand limit does not equal f(2), so f is discontinuous at x = 2.
4. A function that is discontinuous at a point cannot be differentiable there.

In shortf is discontinuous at both x = 1 and x = 2 (the left-hand limit does not match the function's value there), so it is not differentiable at either point.

Watch this explained “The staircase, and the other failure”, 12:04 into Differentiability, and why it forces continuity while the reverse fails