Exercise 5.4 answers: Continuity and Differentiability

Open NCERT p. 130Matches NCERT’s answer

1. This is a quotient, u/v, with u = e^x and v = sin x.
2. Use the quotient rule: dy/dx = (u'v − uv')/v².
3. u' = e^x (the exponential is its own derivative), and v' = cos x.
4. dy/dx = [e^x·sin x − e^x·cos x] / sin²x.
5. Factor e^x out of the numerator: dy/dx = e^x(sin x − cos x)/sin²x.

Answerdy/dx = e^x(sin x − cos x) / sin²x

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Open NCERT p. 130Matches NCERT’s answer

1. This is e raised to a power, and the power is sin⁻¹x.
2. Outer function: e^u. Inner function: u = sin⁻¹x.
3. Derivative of e^u is e^u itself.
4. Derivative of sin⁻¹x is 1/√(1−x²).
5. Multiply the two (chain rule).

Answerdy/dx = e^(sin⁻¹x) / √(1−x²)

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Open NCERT p. 130Matches NCERT’s answer

1. Outer function: e^u. Inner function: u = x³.
2. Derivative of e^u is e^u; derivative of x³ is 3x².
3. Multiply the two.

Answerdy/dx = 3x² e^(x³)

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Open NCERT p. 130Matches NCERT’s answer

1. Three layers, from outside in: sin( ), then tan⁻¹( ), then e⁻ˣ.
2. Derivative of sin(u) is cos(u), with u = tan⁻¹(e⁻ˣ).
3. Derivative of tan⁻¹(v) is 1/(1+v²), with v = e⁻ˣ. Here v² = e⁻²ˣ, so this factor is 1/(1+e⁻²ˣ).
4. Derivative of e⁻ˣ is −e⁻ˣ.
5. Multiply the three factors: dy/dx = cos(tan⁻¹e⁻ˣ) × 1/(1+e⁻²ˣ) × (−e⁻ˣ).
6. To simplify cos(tan⁻¹v), let θ = tan⁻¹v. Then tan θ = v and −π/2 < θ < π/2. So sec²θ = 1 + tan²θ = 1 + v², and cos θ is positive, which gives cos θ = 1/√(1+v²).
7. So cos(tan⁻¹e⁻ˣ) = 1/√(1+e⁻²ˣ). The two denominators combine: √(1+e⁻²ˣ) × (1+e⁻²ˣ) = (1+e⁻²ˣ)^3/2.

Answerdy/dx = −e⁻ˣ cos(tan⁻¹e⁻ˣ)/(1+e⁻²ˣ), which simplifies to −e⁻ˣ / (1+e⁻²ˣ)^3/2

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. 130Matches NCERT’s answer

1. Differentiate: log (cos eˣ).
2. Three layers: log( ), then cos( ), then eˣ.
3. Derivative of log(u) is 1/u, with u = cos(eˣ).
4. Derivative of cos(v) is −sin(v), with v = eˣ.
5. Derivative of eˣ is eˣ.
6. Multiply the three factors and simplify sin/cos to tan.

Answerdy/dx = −eˣ tan(eˣ)

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. 130Matches NCERT’s answer

1. Differentiate term by term; each term is e raised to a power of x.
2. d/dx(eˣ) = eˣ.
3. d/dx(e^x²) = 2x e^x² (chain rule, inner derivative 2x).
4. d/dx(e^x³) = 3x² e^x³.
5. d/dx(e^x⁴) = 4x³ e^x⁴.
6. d/dx(e^x⁵) = 5x⁴ e^x⁵.

Answerdy/dx = eˣ + 2x e^(x²) + 3x² e^(x³) + 4x³ e^(x⁴) + 5x⁴ e^(x⁵)

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Open NCERT p. 130Matches NCERT’s answer

1. A square root of e^√x is the same as e^(√x)/2 — simplify before differentiating.
2. Now it is one exponential: outer e^u, inner u = (√x)/2.
3. Derivative of √x is 1/(2√x), so derivative of u is 1/(4√x).
4. Multiply the two factors.

Answerdy/dx = e^(√x/2) / (4√x), the same as √(e^√x) / (4√x)

Watch this explained “Two worth doing slowly”, 22:12 into Exponential and logarithmic functions, and the two derivatives that make them worth having

Open NCERT p. 130Matches NCERT’s answer

1. Outer function: log(u). Inner function: u = log x.
2. Derivative of log(u) is 1/u; derivative of log x is 1/x.
3. Multiply the two factors.

Answerdy/dx = 1/(x log x)

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Open NCERT p. 130Matches NCERT’s answer

1. This is a quotient: the numerator is u = cos x and the denominator is v = log x.
2. Quotient rule: dy/dx = (v·u′ − u·v′) / v².
3. u′ = −sin x and v′ = 1/x.
4. Substitute: dy/dx = [log x · (−sin x) − cos x · (1/x)] / (log x)².
5. Multiply the top and the bottom by x to clear the 1/x: dy/dx = [−x sin x log x − cos x] / [x (log x)²].

Answerdy/dx = −(x sin x log x + cos x) / (x (log x)²)

Watch this explained “The two derivatives, measured”, 20:22 into Exponential and logarithmic functions, and the two derivatives that make them worth having

Open NCERT p. 130Checked by computer

1. Differentiate: cos (log x + eˣ), x > 0.
2. Outer function: cos(u). Inner function: u = log x + eˣ.
3. Derivative of cos(u) is −sin(u).
4. Derivative of the inner sum: 1/x + eˣ.
5. Multiply the two factors.
6. The answer key at the back of the book prints −1/x + eˣ sin(log x + eˣ); the brackets around 1/x + eˣ have dropped out, so the answer is −(1/x + eˣ) sin(log x + eˣ).

Answerdy/dx = −sin(log x + eˣ) × (1/x + eˣ)

Watch this explained “Two worth doing slowly”, 22:12 into Exponential and logarithmic functions, and the two derivatives that make them worth having