Exercise 5.3 answers: Continuity and Differentiability

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1. Differentiate both sides with respect to x, remembering y is a function of x.
2. Left side: the derivative of 2x is 2, and the derivative of 3y is 3(dy/dx).
3. Right side: the derivative of sin x is cos x.
4. So 2 + 3(dy/dx) = cos x.
5. Solve for dy/dx: dy/dx = (cos x − 2)/3.

Answerdy/dx = (cos x − 2)/3

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x.
2. Left side: 2 + 3(dy/dx).
3. Right side: the derivative of sin y is cos y × (dy/dx), by the chain rule, since y is a function of x.
4. So 2 + 3(dy/dx) = cos y · (dy/dx).
5. Collect the dy/dx terms: 2 = (dy/dx)(cos y − 3).
6. dy/dx = 2/(cos y − 3).

Answerdy/dx = 2/(cos y − 3)

Watch this explained “Collected, and solved”, 12:47 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x; a and b are constants.
2. Left side: the derivative of ax is a, and the derivative of by² is 2by(dy/dx), by the chain rule.
3. Right side: the derivative of cos y is −sin y · (dy/dx).
4. So a + 2by(dy/dx) = −sin y · (dy/dx).
5. Collect: a = −(dy/dx)(sin y + 2by).
6. dy/dx = −a / (sin y + 2by).

Answerdy/dx = −a / (sin y + 2by)

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x.
2. Left side: xy needs the product rule, giving y + x(dy/dx); y² gives 2y(dy/dx) by the chain rule.
3. Right side: the derivative of tan x is sec²x, and the derivative of y is dy/dx.
4. So y + x(dy/dx) + 2y(dy/dx) = sec²x + dy/dx.
5. Collect the dy/dx terms: (dy/dx)(x + 2y − 1) = sec²x − y.
6. dy/dx = (sec²x − y) / (x + 2y − 1).

Answerdy/dx = (sec²x − y) / (x + 2y − 1)

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x; the right side, 100, differentiates to 0.
2. Left side: x² gives 2x; xy needs the product rule, giving y + x(dy/dx); y² gives 2y(dy/dx).
3. So 2x + y + x(dy/dx) + 2y(dy/dx) = 0.
4. Collect: (dy/dx)(x + 2y) = −(2x + y).
5. dy/dx = −(2x + y) / (x + 2y).

Answerdy/dx = −(2x + y) / (x + 2y)

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x; 81 differentiates to 0.
2. x³ gives 3x². x²y needs the product rule: 2xy + x²(dy/dx). xy² needs the product rule: y² + 2xy(dy/dx). y³ gives 3y²(dy/dx).
3. Add them up: 3x² + 2xy + x²(dy/dx) + y² + 2xy(dy/dx) + 3y²(dy/dx) = 0.
4. Collect the dy/dx terms: (dy/dx)(x² + 2xy + 3y²) = −(3x² + 2xy + y²).
5. dy/dx = −(3x² + 2xy + y²) / (x² + 2xy + 3y²).

Answerdy/dx = −(3x² + 2xy + y²) / (x² + 2xy + 3y²)

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. κ is a constant, so the right side differentiates to 0.
2. sin²y differentiates, by the chain rule, to 2 sin y cos y (dy/dx), which is sin(2y)(dy/dx).
3. cos(xy) differentiates, by the chain rule with the product rule inside xy, to −sin(xy) × [y + x(dy/dx)].
4. So sin(2y)(dy/dx) − sin(xy)[y + x(dy/dx)] = 0.
5. Expand: sin(2y)(dy/dx) − y·sin(xy) − x·sin(xy)(dy/dx) = 0.
6. Collect: (dy/dx)[sin(2y) − x·sin(xy)] = y·sin(xy).
7. dy/dx = y·sin(xy) / [sin(2y) − x·sin(xy)].

Answerdy/dx = y·sin(xy) / [sin(2y) − x·sin(xy)]

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Differentiate both sides with respect to x; 1 differentiates to 0.
2. sin²x differentiates to 2 sin x cos x, which is sin 2x.
3. cos²y differentiates, by the chain rule, to −2 cos y sin y (dy/dx), which is −sin(2y)(dy/dx).
4. So sin 2x − sin(2y)(dy/dx) = 0.
5. dy/dx = sin 2x / sin 2y.

Answerdy/dx = sin 2x / sin 2y

Watch this explained “The one new permission”, 11:07 into The chain rule, and differentiating a relation without first solving it for y

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1. Substitute x = tan θ, so θ = tan⁻¹x. Then 2x/(1 + x²) = 2 tan θ/(1 + tan²θ) = sin 2θ.
2. sin⁻¹(sin 2θ) equals 2θ only when 2θ lies between −π/2 and π/2, that is when −1 < x < 1. The question prints no interval, so we work on this range, which is the one the standard answer assumes.
3. On that range y = 2θ = 2 tan⁻¹x.
4. Differentiate: dy/dx = 2 × 1/(1 + x²) = 2/(1 + x²).
5. For x > 1 or x < −1 the same substitution gives y = π − 2 tan⁻¹x or y = −π − 2 tan⁻¹x, so there dy/dx = −2/(1 + x²) instead. At x = ±1 there is no derivative.

Answerdy/dx = 2/(1 + x²), for −1 < x < 1

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Substitute x = tan θ. Then (3x − x³)/(1 − 3x²) becomes tan 3θ, a standard identity.
2. So y = tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹x, valid on the given interval.
3. Differentiate: dy/dx = 3 × 1/(1 + x²).

Answerdy/dx = 3/(1 + x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Substitute x = tan θ. Then (1 − x²)/(1 + x²) becomes cos 2θ, a standard identity.
2. So y = cos⁻¹(cos 2θ) = 2θ = 2 tan⁻¹x, valid on the given interval.
3. Differentiate: dy/dx = 2 × 1/(1 + x²).

Answerdy/dx = 2/(1 + x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Substitute x = tan θ. Then (1 − x²)/(1 + x²) becomes cos 2θ, which equals sin(π/2 − 2θ).
2. So y = sin⁻¹(sin(π/2 − 2θ)) = π/2 − 2θ = π/2 − 2 tan⁻¹x, valid on the given interval.
3. Differentiate: dy/dx = −2 × 1/(1 + x²).

Answerdy/dx = −2/(1 + x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Substitute x = tan θ. Then 2x/(1 + x²) becomes sin 2θ, which equals cos(π/2 − 2θ).
2. So y = cos⁻¹(cos(π/2 − 2θ)) = π/2 − 2θ = π/2 − 2 tan⁻¹x, valid on the given interval.
3. Differentiate: dy/dx = −2 × 1/(1 + x²).

Answerdy/dx = −2/(1 + x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Substitute x = sin θ. Then 2x√(1 − x²) becomes 2 sin θ cos θ = sin 2θ, a standard identity.
2. So y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x, valid on the given interval.
3. Differentiate: dy/dx = 2 × 1/√(1 − x²).

Answerdy/dx = 2/√(1 − x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions

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1. Rewrite the secant inverse as a cosine inverse: sec⁻¹(1/(2x² − 1)) = cos⁻¹(2x² − 1).
2. Substitute x = cos θ. Then 2x² − 1 becomes cos 2θ, a standard identity.
3. So y = cos⁻¹(cos 2θ) = 2θ = 2 cos⁻¹x, valid on the given interval.
4. Differentiate: dy/dx = 2 × (−1/√(1 − x²)) = −2/√(1 − x²).

Answerdy/dx = −2/√(1 − x²)

Watch this explained “Substitute first, differentiate second”, 11:55 into Getting the derivatives of the inverse trigonometric functions