Exercise 5.7 answers: Continuity and Differentiability

Open NCERT p. 139Matches NCERT’s answer

1. First derivative: dy/dx = 2x + 3.
2. Differentiate again: d²y/dx² = 2.

Answerd²y/dx² = 2.

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. First derivative: dy/dx = 20x¹⁹.
2. Differentiate again: d²y/dx² = 20 × 19 × x¹⁸ = 380x¹⁸.

Answerd²y/dx² = 380x¹⁸.

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = x cos x is a product of x and cos x, so use the product rule.
2. dy/dx = (1)(cos x) + (x)(−sin x) = cos x − x sin x.
3. Differentiate again, term by term.
4. d/dx(cos x) = −sin x.
5. d/dx(x sin x) = sin x + x cos x, by the product rule again.
6. d²y/dx² = −sin x − (sin x + x cos x) = −2 sin x − x cos x.

Answerd²y/dx² = −x cos x − 2 sin x

Watch this explained “Which four need the product rule”, 10:44 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = log x, so dy/dx = 1/x.
2. Write 1/x as x⁻¹ and differentiate again using the power rule.
3. d²y/dx² = −1·x⁻² = −1/x².

Answerd²y/dx² = −1/x²

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = x³ log x is a product, so use the product rule.
2. dy/dx = 3x² log x + x³·(1/x) = 3x² log x + x².
3. Differentiate again, term by term.
4. d/dx(3x² log x) = 6x log x + 3x²·(1/x) = 6x log x + 3x.
5. d/dx(x²) = 2x.
6. d²y/dx² = 6x log x + 3x + 2x = 6x log x + 5x.

Answerd²y/dx² = 6x log x + 5x

Watch this explained “Which four need the product rule”, 10:44 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = eˣ sin 5x is a product, so use the product rule.
2. dy/dx = eˣ sin 5x + eˣ·5 cos 5x = eˣ(sin 5x + 5 cos 5x).
3. Differentiate again using the product rule on eˣ and (sin 5x + 5 cos 5x).
4. d²y/dx² = eˣ(sin 5x + 5 cos 5x) + eˣ(5 cos 5x − 25 sin 5x).
5. Collect like terms: d²y/dx² = eˣ(10 cos 5x − 24 sin 5x).

Answerd²y/dx² = eˣ(10 cos 5x − 24 sin 5x)

Watch this explained “Which four need the product rule”, 10:44 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = e⁶ˣ cos 3x is a product, so use the product rule.
2. dy/dx = 6e⁶ˣ cos 3x − 3e⁶ˣ sin 3x = e⁶ˣ(6 cos 3x − 3 sin 3x).
3. Differentiate again using the product rule on e⁶ˣ and (6 cos 3x − 3 sin 3x).
4. d²y/dx² = 6e⁶ˣ(6 cos 3x − 3 sin 3x) + e⁶ˣ(−18 sin 3x − 9 cos 3x).
5. Collect like terms: d²y/dx² = e⁶ˣ(27 cos 3x − 36 sin 3x).

Answerd²y/dx² = e⁶ˣ(27 cos 3x − 36 sin 3x)

Watch this explained “Which four need the product rule”, 10:44 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = tan⁻¹x, so dy/dx = 1/(1 + x²).
2. Write this as (1 + x²)⁻¹ and use the chain rule to differentiate again.
3. d²y/dx² = −1·(1 + x²)⁻²·2x = −2x/(1 + x²)².

Answerd²y/dx² = −2x/(1 + x²)²

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = log(log x). By the chain rule, dy/dx = (1/log x)·(1/x) = 1/(x log x).
2. Write this as (x log x)⁻¹ and use the chain rule again.
3. d/dx(x log x) = log x + 1, by the product rule.
4. d²y/dx² = −(log x + 1)/(x log x)².

Answerd²y/dx² = −(log x + 1)/(x² (log x)²)

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139Matches NCERT’s answer

1. y = sin(log x). By the chain rule, dy/dx = cos(log x)·(1/x) = cos(log x)/x.
2. Differentiate again using the quotient (or product) rule on cos(log x) and 1/x.
3. d/dx[cos(log x)] = −sin(log x)/x, by the chain rule.
4. d²y/dx² = [−sin(log x)/x]·(1/x) + cos(log x)·(−1/x²).
5. d²y/dx² = −sin(log x)/x² − cos(log x)/x² = −[sin(log x) + cos(log x)]/x².

Answerd²y/dx² = −[sin(log x) + cos(log x)]/x²

Watch this explained “Ten routine second derivatives”, 8:45 into Differentiating twice, and what the second derivative is for

Open NCERT p. 139One way to think about it

1. Differentiate y = 5 cos x − 3 sin x once.
2. dy/dx = −5 sin x − 3 cos x.
3. Differentiate again.
4. d²y/dx² = −5 cos x + 3 sin x = −(5 cos x − 3 sin x).
5. So d²y/dx² = −y, which means d²y/dx² + y = 0.

In shortd²y/dx² + y = 0, as proved above.

Watch this explained “The task that changes without warning”, 12:32 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140Matches NCERT’s answer

1. y = cos⁻¹x, so dy/dx = −1/√(1 − x²).
2. Differentiate again: d²y/dx² = −x/(1 − x²)^(3/2).
3. Since y = cos⁻¹x, we have x = cos y, and 1 − x² = 1 − cos²y = sin²y.
4. So (1 − x²)^(3/2) = sin³y (as 0 < y < π means sin y > 0).
5. Substitute x = cos y: d²y/dx² = −cos y / sin³y.

Answerd²y/dx² = −cos y / sin³y

Watch this explained “An answer in the output”, 21:31 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140One way to think about it

1. Let t = log x. Differentiate y = 3 cos t + 4 sin t using the chain rule.
2. y1 = dy/dx = −3 sin(log x)·(1/x) + 4 cos(log x)·(1/x) = (1/x)[4 cos(log x) − 3 sin(log x)].
3. Multiply both sides by x: x·y1 = 4 cos(log x) − 3 sin(log x).
4. Differentiate both sides again with respect to x.
5. On the left, use the product rule: d/dx(x·y1) = y1 + x·y2.
6. On the right: d/dx[4 cos(log x) − 3 sin(log x)] = [−4 sin(log x) − 3 cos(log x)]·(1/x).
7. So y1 + x·y2 = −(1/x)[3 cos(log x) + 4 sin(log x)] = −y/x.
8. Multiply through by x: x·y1 + x²·y2 = −y.
9. So x² y2 + x y1 + y = 0.

In shortx² y2 + x y1 + y = 0, as proved above.

Watch this explained “The task that changes without warning”, 12:32 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140One way to think about it

1. Given: y = Ae^(mx) + Be^(nx). To show: d²y/dx² − (m + n) dy/dx + mny = 0.
2. Differentiate y = Ae^(mx) + Be^(nx) once.
3. y1 = Am·e^(mx) + Bn·e^(nx).
4. Differentiate again.
5. y2 = Am²·e^(mx) + Bn²·e^(nx).
6. Now compute y2 − (m + n)y1 + mny, collecting the e^(mx) terms and the e^(nx) terms separately.
7. e^(mx) terms: A[m² − (m+n)m + mn] = A[m² − m² − mn + mn] = 0.
8. e^(nx) terms: B[n² − (m+n)n + mn] = B[n² − mn − n² + mn] = 0.
9. Both groups are zero, so y2 − (m + n)y1 + mny = 0.

In shortd²y/dx² − (m + n) dy/dx + mny = 0, as proved above.

Watch this explained “The column that makes it mechanical”, 14:05 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140One way to think about it

1. Differentiate y = 500e^(7x) + 600e^(−7x) once.
2. y1 = 3500e^(7x) − 4200e^(−7x).
3. Differentiate again.
4. y2 = 24500e^(7x) + 29400e^(−7x).
5. Factor out 49: y2 = 49(500e^(7x) + 600e^(−7x)) = 49y.

In shortd²y/dx² = 49y, as proved above.

Watch this explained “The column that makes it mechanical”, 14:05 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140One way to think about it

1. Given: eʸ(x + 1) = 1. To show: d²y/dx² = (dy/dx)².
2. From e^y(x + 1) = 1, we get e^y = 1/(x + 1), so y = −log(x + 1).
3. Differentiate: y1 = dy/dx = −1/(x + 1).
4. Differentiate again: y2 = d²y/dx² = 1/(x + 1)².
5. Now (y1)² = [−1/(x+1)]² = 1/(x+1)².
6. So y2 = (y1)², that is d²y/dx² = (dy/dx)².

In shortd²y/dx² = (dy/dx)², as proved above.

Watch this explained “The task that changes without warning”, 12:32 into Differentiating twice, and what the second derivative is for

Open NCERT p. 140One way to think about it

1. Let t = tan⁻¹x, so y = t². Differentiate using the chain rule.
2. y1 = 2t·(1/(1 + x²)) = 2 tan⁻¹x / (1 + x²).
3. Multiply both sides by (1 + x²): (1 + x²)y1 = 2 tan⁻¹x.
4. Differentiate both sides again with respect to x.
5. Left side, by the product rule: (1 + x²)y2 + 2x·y1.
6. Right side: 2·(1/(1 + x²)).
7. So (1 + x²)y2 + 2x·y1 = 2/(1 + x²).
8. Multiply through by (1 + x²): (1 + x²)² y2 + 2x(1 + x²) y1 = 2.

In short(x² + 1)² y2 + 2x(x² + 1) y1 = 2, as proved above.

Watch this explained “Clearing the radical”, 15:25 into Differentiating twice, and what the second derivative is for