Miscellaneous Exercise answers: Continuity and Differentiability

Open NCERT p. 144Matches NCERT’s answer

1. y = (3x² − 9x + 5)⁹ is a function raised to a power, so use the chain rule.
2. Outer rule: differentiate (something)⁹ to get 9(something)⁸.
3. Inner rule: differentiate 3x² − 9x + 5 to get 6x − 9.
4. dy/dx = 9(3x² − 9x + 5)⁸ · (6x − 9).
5. Take out the common factor 3 from (6x − 9): dy/dx = 27(2x − 3)(3x² − 9x + 5)⁸.

Answerdy/dx = 27(2x − 3)(3x² − 9x + 5)⁸

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. 144Matches NCERT’s answer

1. y = sin³x + cos⁶x. Differentiate each term separately using the chain rule.
2. d/dx(sin³x) = 3 sin²x · cos x (outer power rule, inner derivative of sin x is cos x).
3. d/dx(cos⁶x) = 6 cos⁵x · (−sin x) = −6 sin x cos⁵x.
4. Add the two results: dy/dx = 3 sin²x cos x − 6 sin x cos⁵x.
5. Take out the common factor 3 sin x cos x: dy/dx = 3 sin x cos x (sin x − 2 cos⁴x).

Answerdy/dx = 3 sin x cos x (sin x − 2 cos⁴x)

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

Open NCERT p. 144Matches NCERT’s answer

1. y = (5x)^(3 cos 2x) has both a variable base and a variable exponent, so take logarithms of both sides first.
2. log y = 3 cos 2x · log(5x).
3. Differentiate both sides with respect to x, using the product rule on the right.
4. (1/y)·dy/dx = 3 cos 2x · (1/x) + log(5x) · (−6 sin 2x).
5. Multiply both sides by y: dy/dx = y[3 cos 2x / x − 6 log(5x) sin 2x].
6. Substitute back y = (5x)^(3 cos 2x): dy/dx = (5x)^(3 cos 2x) [3 cos 2x / x − 6 log(5x) sin 2x].

Answerdy/dx = (5x)^(3 cos 2x) [3 cos 2x / x − 6 log(5x) sin 2x]

Watch this explained “The formula, term by term”, 3:28 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 144Matches NCERT’s answer

1. Differentiate: sin⁻¹(x√x), 0 ≤ x ≤ 1.
2. y = sin⁻¹(x√x). Note that x√x = x·x^(1/2) = x^(3/2), so y = sin⁻¹(x^(3/2)).
3. Use the chain rule: derivative of sin⁻¹u is 1/√(1 − u²), times du/dx.
4. Here u = x^(3/2), so u² = x³, and du/dx = (3/2)x^(1/2).
5. dy/dx = [1/√(1 − x³)] · (3/2)√x.
6. dy/dx = 3√x / [2√(1 − x³)].

Answerdy/dx = 3√x / [2√(1 − x³)]

Watch this explained “Two items worth doing slowly”, 14:23 into Getting the derivatives of the inverse trigonometric functions

Open NCERT p. 144Matches NCERT’s answer

1. Differentiate: cos⁻¹(x/2)/√(2x + 7), −2 < x < 2.
2. y = cos⁻¹(x/2) / (2x + 7)^(1/2) is a quotient, so use the quotient rule.
3. Let u = cos⁻¹(x/2), so du/dx = −(1/2) / √(1 − x²/4) = −1/√(4 − x²).
4. Let v = (2x + 7)^(1/2), so dv/dx = (1/2)(2x + 7)^(−1/2)·2 = 1/√(2x + 7).
5. Quotient rule: dy/dx = [v·(du/dx) − u·(dv/dx)] / v².
6. dy/dx = [√(2x+7)·(−1/√(4−x²)) − cos⁻¹(x/2)·(1/√(2x+7))] / (2x + 7).
7. Multiply numerator and denominator by √(2x+7)·√(4−x²) to clear the inner square roots.
8. dy/dx = −[(2x + 7) + √(4 − x²)·cos⁻¹(x/2)] / [√(4 − x²)·(2x + 7)^(3/2)].

Answerdy/dx = −[2x + 7 + √(4 − x²)·cos⁻¹(x/2)] / [√(4 − x²)·(2x + 7)^(3/2)]

Watch the lesson Getting the derivatives of the inverse trigonometric functions

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1. Differentiate: cot⁻¹[(√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x))], 0 < x < π/2.
2. Write 1 + sin x and 1 − sin x as perfect squares using half angles: 1 + sin x = (cos(x/2) + sin(x/2))² and 1 − sin x = (cos(x/2) − sin(x/2))².
3. Since 0 < x/2 < π/4 here, cos(x/2) is bigger than sin(x/2), so both square roots come out positive without any sign trouble: √(1+sin x) = cos(x/2)+sin(x/2) and √(1−sin x) = cos(x/2)−sin(x/2).
4. Add these two: the top of the fraction becomes 2cos(x/2). Subtract them: the bottom becomes 2sin(x/2).
5. So the fraction inside cot⁻¹ simplifies to cos(x/2)/sin(x/2), which is cot(x/2).
6. y = cot⁻¹(cot(x/2)) = x/2, because x/2 already lies between 0 and π/2, the range where cot⁻¹ just gives back its own angle.
7. Differentiate the simple function y = x/2 to get dy/dx.

Answerdy/dx = 1/2

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

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1. Let y = (log x)^(log x). Both the base and the power contain x, so take the log of both sides first: log y = log x · log(log x).
2. Differentiate both sides with respect to x. On the left, the chain rule gives (1/y)(dy/dx).
3. On the right, use the product rule on log x · log(log x): its derivative is (1/x)·log(log x) + log x · (1/log x) · (1/x), which simplifies to (1/x)[log(log x) + 1].
4. So (1/y)(dy/dx) = [1 + log(log x)]/x.
5. Multiply both sides by y to get dy/dx = (log x)^(log x) · [1 + log(log x)]/x.

Answerdy/dx = (log x)^(log x) · [1 + log(log x)]/x

Watch this explained “Taking logarithms of both sides”, 1:02 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 144Matches NCERT’s answer

1. Let u = a cos x + b sin x, so y = cos u is a function of a function.
2. Differentiate the outer function: d(cos u)/du = −sin u.
3. Differentiate the inner function: du/dx = −a sin x + b cos x.
4. By the chain rule, dy/dx = (−sin u) × (−a sin x + b cos x) = (a sin x − b cos x) sin(a cos x + b sin x).

Answerdy/dx = (a sin x − b cos x) sin(a cos x + b sin 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. 144Matches NCERT’s answer

1. Let u = sin x − cos x, which is positive for every x in this interval. Then y = u^u.
2. Take the log of both sides: log y = u log u.
3. Differentiate with respect to x: (1/y)(dy/dx) = u′(log u + 1), where u′ = cos x + sin x.
4. Multiply through by y to get dy/dx = (sin x − cos x)^(sin x − cos x) · (cos x + sin x) · [1 + log(sin x − cos x)].

Answerdy/dx = (sin x − cos x)^(sin x − cos x) · (cos x + sin x) · [1 + log(sin x − cos x)]

Watch this explained “Taking logarithms of both sides”, 1:02 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 144Matches NCERT’s answer

1. Differentiate: xˣ + xᵃ + aˣ + aᵃ, for some fixed a > 0 and x > 0.
2. Differentiate the four terms one by one, because each one needs a different rule.
3. x^x has a variable base AND a variable power, so use logarithmic differentiation: take log y₁ = x log x, differentiate to get y₁′/y₁ = 1 + log x, so d/dx(x^x) = x^x(1+log x).
4. x^a has a variable base but a FIXED power a, so this is the ordinary power rule: d/dx(x^a) = a·x^(a−1).
5. a^x has a FIXED base but a variable power, so this is the ordinary exponential rule: d/dx(a^x) = a^x · log a.
6. a^a has both parts fixed, so it is just a constant number — its derivative is 0.
7. Add the four pieces: dy/dx = x^x(1+log x) + a·x^(a−1) + a^x·log a.

Answerdy/dx = x^x(1 + log x) + a·x^(a−1) + a^x·log a

Watch this explained “The shape no rule reaches”, 0:00 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 145Matches NCERT’s answer

1. Both terms have a variable base and a variable exponent, so use logarithmic differentiation on each one separately.
2. Let u = x^(x²−3). log u = (x²−3) log x. Differentiate: u′/u = 2x log x + (x²−3)/x, so u′ = x^(x²−3)[2x log x + (x²−3)/x].
3. Let v = (x−3)^(x²). log v = x² log(x−3). Differentiate: v′/v = 2x log(x−3) + x²/(x−3), so v′ = (x−3)^(x²)[2x log(x−3) + x²/(x−3)].
4. dy/dx is just u′ + v′, since y = u + v.

Answerdy/dx = x^(x²−3)[2x log x + (x²−3)/x] + (x−3)^(x²)[2x log(x−3) + x²/(x−3)]

Watch this explained “The formula, term by term”, 3:28 into Taking logarithms first, when the base and the power both vary

Open NCERT p. 145Matches NCERT’s answer

1. Neither x nor y is written directly in terms of the other — both are written in terms of t. So differentiate each one with respect to t first.
2. dy/dt = 12 sin t.
3. dx/dt = 10(1 − cos t).
4. dy/dx = (dy/dt)/(dx/dt) = 12 sin t / [10(1 − cos t)] = 6 sin t / [5(1 − cos t)].

Answerdy/dx = 6 sin t / [5(1 − cos t)]

Watch this explained “The cycloid, and a half angle”, 11:08 into Curves given through a parameter, and the derivative recovered by the chain rule

Open NCERT p. 145Matches NCERT’s answer

1. Differentiate sin⁻¹x on its own: its derivative is 1/√(1−x²).
2. Differentiate sin⁻¹√(1−x²) using the chain rule with u = √(1−x²): the derivative of sin⁻¹u is 1/√(1−u²), and here 1−u² = x², so 1/√(1−u²) = 1/x (using x > 0).
3. Multiply by du/dx = −x/√(1−x²): this term's derivative is (1/x) × (−x/√(1−x²)) = −1/√(1−x²).
4. Add the two derivatives: 1/√(1−x²) + (−1/√(1−x²)) = 0.

Answerdy/dx = 0

Watch this explained “A sum that does not move”, 15:25 into Getting the derivatives of the inverse trigonometric functions

Open NCERT p. 145One way to think about it

1. Move one term across: x√(1 + y) = −y√(1 + x).
2. Square both sides to remove the square roots: x²(1 + y) = y²(1 + x).
3. Expand and bring everything to one side: x² + x²y − y² − xy² = 0.
4. Group in pairs: (x² − y²) + xy(x − y) = 0, that is (x − y)(x + y) + xy(x − y) = 0, so (x − y)(x + y + xy) = 0.
5. The factor x − y = 0 cannot be the curve: if y = x, the given equation becomes 2x√(1 + x) = 0, which is true only at x = 0, not for every x with −1 < x < 1. So x + y + xy = 0.
6. Solve for y: y(1 + x) = −x, so y = −x/(1 + x). Check it in the original equation: 1 + y = 1/(1 + x), so x√(1 + y) + y√(1 + x) = x/√(1 + x) − x/√(1 + x) = 0.
7. Differentiate by the quotient rule: dy/dx = [−(1 + x) − (−x)(1)]/(1 + x)² = −1/(1 + x)².

In shortHence dy/dx = −1/(1 + x)², as required.

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

Open NCERT p. 145One way to think about it

1. Differentiate (x − a)² + (y − b)² = c² with respect to x: 2(x − a) + 2(y − b)·dy/dx = 0, so dy/dx = −(x − a)/(y − b). (This needs y ≠ b.)
2. Differentiate (x − a) + (y − b)·dy/dx = 0 once more: 1 + (dy/dx)² + (y − b)·d²y/dx² = 0, so d²y/dx² = −[1 + (dy/dx)²]/(y − b).
3. Use the circle's equation: 1 + (dy/dx)² = 1 + (x − a)²/(y − b)² = [(x − a)² + (y − b)²]/(y − b)² = c²/(y − b)².
4. So [1 + (dy/dx)²]^(3/2) = c³/|y − b|³, and d²y/dx² = −c²/(y − b)³.
5. Divide: [1 + (dy/dx)²]^(3/2) ÷ d²y/dx² = (c³/|y − b|³) × (−(y − b)³/c²) = −c·(y − b)³/|y − b|³.
6. (y − b)³/|y − b|³ is 1 when y > b (the upper half of the circle) and −1 when y < b (the lower half). So the expression is −c on the upper half and c on the lower half.
7. Either way its value is fixed by the radius c alone and does not involve a or b, so it is a constant independent of a and b.

In shortThe expression equals −c on the upper half of the circle (y > b) and c on the lower half (y < b), so it is a constant independent of a and b.

Watch this explained “The circle, and its radius”, 29:23 into Differentiating twice, and what the second derivative is for

Open NCERT p. 145One way to think about it

1. Differentiate both sides with respect to x. The left side gives −sin y · dy/dx.
2. The right side is a product of x and cos(a+y), so use the product rule: cos(a+y) + x·(−sin(a+y))·dy/dx.
3. Set them equal: −sin y·dy/dx = cos(a+y) − x sin(a+y)·dy/dx.
4. Collect all the dy/dx terms on one side: dy/dx·[x sin(a+y) − sin y] = cos(a+y), so dy/dx = cos(a+y) / [x sin(a+y) − sin y].
5. Now replace x using the original relation, x = cos y / cos(a+y): the denominator becomes [cos y·sin(a+y) − sin y·cos(a+y)] / cos(a+y).
6. The numerator of that bracket is sin[(a+y) − y] = sin a, by the sine-difference formula. So the denominator simplifies to sin a / cos(a+y).
7. So dy/dx = cos(a+y) ÷ [sin a/cos(a+y)] = cos²(a+y)/sin a, which is exactly what was to be proved.

In shortHence dy/dx = cos²(a+y)/sin a, as required.

Watch this explained “A relation carrying a fixed angle”, 16:38 into The chain rule, and differentiating a relation without first solving it for y

Open NCERT p. 145Matches NCERT’s answer

1. Differentiate x with respect to t: dx/dt = a(−sin t + sin t + t cos t) = a t cos t.
2. Differentiate y with respect to t: dy/dt = a(cos t − cos t + t sin t) = a t sin t.
3. So dy/dx = (dy/dt)/(dx/dt) = (a t sin t)/(a t cos t) = tan t.
4. For the second derivative, differentiate dy/dx = tan t with respect to t, then divide by dx/dt once more: d/dt(tan t) = sec²t.
5. d²y/dx² = sec²t ÷ (a t cos t) = 1/(a t cos³t), since sec²t/cos t = 1/cos³t.

Answerd²y/dx² = 1/(a t cos³t) (equivalently sec³t / (a t))

Watch this explained “Divide twice, do not differentiate twice”, 23:43 into Differentiating twice, and what the second derivative is for

Open NCERT p. 145Checked by computer

1. Write f(x) = |x|³ without the modulus. For x ≥ 0, |x| = x, so f(x) = x³. For x < 0, |x| = −x, so f(x) = −x³.
2. Differentiate each piece: for x > 0, f′(x) = 3x²; for x < 0, f′(x) = −3x².
3. At x = 0 use the definition: [f(h) − f(0)]/h = |h|³/h = h|h|, which tends to 0 whether h → 0 from the left or the right. So f′(0) = 0, and f′(x) = 3x|x| for every real x.
4. Differentiate again away from 0: for x > 0, f″(x) = 6x; for x < 0, f″(x) = −6x.
5. At x = 0 use the definition on f′: [f′(h) − f′(0)]/h = 3h|h|/h = 3|h|, which tends to 0 from both sides. So f″(0) exists and equals 0.
6. So f″(x) exists for every real x. It is 6x for x ≥ 0 and −6x for x < 0, which is 6|x| in one formula.

AnswerYes, f″(x) exists for every real x, and f″(x) = 6|x|.

Watch this explained “A ladder of joins”, 6:32 into Differentiating twice, and what the second derivative is for

Open NCERT p. 145Checked by computer

1. Start from the given identity sin(A+B) = sin A cos B + cos A sin B, treating B as fixed and A as the variable.
2. Differentiate both sides with respect to A. The left side, by the chain rule, gives cos(A+B).
3. The right side, term by term, gives cos A cos B − sin A sin B (since B is held fixed, cos B and sin B behave like constants).
4. Since the two sides were equal before differentiating and each side was differentiated the same way, the results must be equal: cos(A+B) = cos A cos B − sin A sin B.

Answercos(A + B) = cos A cos B − sin A sin B

Open NCERT p. 145Checked by computerAnswers can differ: one example

1. Try f(x) = |x−1| + |x−2|, adding two shifted modulus functions.
2. Each modulus function |x−1| and |x−2| is continuous everywhere (a modulus never has a break), so their sum is continuous everywhere too.
3. Away from x = 1 and x = 2, both modulus pieces are just straight lines locally, so f is a straight line on each of the three stretches (x < 1, 1 < x < 2, x > 2), and a straight line is differentiable everywhere on its stretch.
4. At x = 1: the left-hand slope and right-hand slope of |x−1| disagree there (it has a corner), while |x−2| is smooth there, so the sum also has a corner — f is not differentiable at x = 1.
5. At x = 2: by the same reasoning with the roles swapped, f is not differentiable at x = 2 either.
6. Nowhere else does either modulus piece have a corner, so f is differentiable at every other point.
7. So f(x) = |x−1| + |x−2| is continuous everywhere and fails to be differentiable at exactly the two points x = 1 and x = 2.

AnswerYes. For example, f(x) = |x−1| + |x−2| is continuous everywhere but is not differentiable at exactly the two points x = 1 and x = 2.

Watch this explained “Continuous everywhere, failing exactly twice”, 13:54 into Differentiability, and why it forces continuity while the reverse fails

Open NCERT p. 145One way to think about it

1. Expand the determinant along the first row, since that is the only row containing x: y = f(x)(mc − nb) − g(x)(lc − na) + h(x)(lb − ma).
2. Notice that (mc − nb), (lc − na) and (lb − ma) are all built from the constants l, m, n, a, b, c only — none of them contain x, so each is just a fixed number.
3. Differentiate y term by term, treating those bracket quantities as constants: dy/dx = f′(x)(mc − nb) − g′(x)(lc − na) + h′(x)(lb − ma).
4. This is exactly the expansion, along the first row, of the determinant with f′(x), g′(x), h′(x) in place of f(x), g(x), h(x), and the same second and third rows.
5. So dy/dx = |f′(x) g′(x) h′(x); l m n; a b c|, as required.

In shortHence dy/dx = |f′(x) g′(x) h′(x); l m n; a b c|, as required.

Open NCERT p. 145One way to think about it

1. Given: y = e^(a cos⁻¹ x), −1 ≤ x ≤ 1. To show: (1 − x²) d²y/dx² − x dy/dx − a²y = 0.
2. Differentiate y = e^(a cos⁻¹x) by the chain rule: dy/dx = e^(a cos⁻¹x) · a · (−1/√(1 − x²)) = −a y/√(1 − x²). This holds for −1 < x < 1, where √(1 − x²) is not 0.
3. Clear the square root from the denominator: √(1 − x²)·dy/dx = −a y.
4. Differentiate both sides with respect to x, using the product rule on the left: √(1 − x²)·d²y/dx² + (−x/√(1 − x²))·dy/dx = −a·dy/dx.
5. Multiply every term by √(1 − x²): (1 − x²)·d²y/dx² − x·dy/dx = −a·√(1 − x²)·dy/dx.
6. By the second line, √(1 − x²)·dy/dx = −a y, so the right side is −a·(−a y) = a²y.
7. So (1 − x²)·d²y/dx² − x·dy/dx = a²y, that is (1 − x²)·d²y/dx² − x·dy/dx − a²y = 0.

In shortHence (1 − x²)·d²y/dx² − x·dy/dx − a²y = 0, as required.

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