Notation: y′ = dy/dx, y₁ = dy/dx, y₂ = d²y/dx², LHL/RHL = left/right hand limit, LHD/RHD = left/right hand derivative, log means natural log, [x] is the greatest integer function.
Exercise 5.1
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = –3 and at x = 5.
Answer
f is a polynomial, so lim(x
⇒ c) f(x) = 5c – 3 = f(c) for every c.
At x = 0: lim = –3 = f(0). At x = –3: lim = –18 = f(–3). At x = 5: lim = 22 = f(5). Hence f is continuous at all three points.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Answer
lim(x
⇒ 3) (2x² – 1) = 2(9) – 1 = 17 and f(3) = 17. Limit = value, so f is continuous at x = 3.
Examine the following functions for continuity. (a) f(x) = x – 5 (b) f(x) = 1/(x – 5), x ≠ 5 (c) f(x) = (x² – 25)/(x + 5), x ≠ –5 (d) f(x) = |x – 5|
Answer
(a) For any real c, lim(x
⇒ c) (x – 5) = c – 5 = f(c). Continuous everywhere.
(b) Domain is R – {5}. For c ≠ 5, lim(x
⇒ c) 1/(x – 5) = 1/(c – 5) = f(c). Continuous at every point of its domain, so it is a continuous function.
(c) For x ≠ –5, f(x) = (x – 5)(x + 5)/(x + 5) = x – 5. For every c ≠ –5, lim = c – 5 = f(c). Continuous on its domain.
(d) f(x) = 5 – x for x < 5 and x – 5 for x ≥ 5. Both pieces are polynomials, so f is continuous at every c ≠ 5. At x = 5: LHL = lim(5 – x) = 0, RHL = lim(x – 5) = 0, f(5) = 0. So f is continuous everywhere.
Prove that the function f(x) = xⁿ is continuous at x = n, where n is a positive integer.
Answer
f(x) = xⁿ is a polynomial. lim(x
⇒ n) xⁿ = nⁿ = f(n). Hence f is continuous at x = n.
Is the function f defined by f(x) = x, if x ≤ 1; 5, if x > 1 continuous at x = 0? At x = 1? At x = 2?
Answer
At x = 0: near 0, f(x) = x, so lim = 0 = f(0)
⇒ continuous.
At x = 1: LHL = lim(x
⇒ 1⁻) x = 1, RHL = lim(x
⇒ 1⁺) 5 = 5. LHL ≠ RHL
⇒ not continuous at x = 1.
At x = 2: near 2, f(x) = 5, so lim = 5 = f(2)
⇒ continuous.
Find all points of discontinuity of f, where f is defined by (Exercises 6 to 12):
f(x) = 2x + 3, if x ≤ 2; 2x – 3, if x > 2
Answer
For x < 2 and x > 2, f is a polynomial, hence continuous. At x = 2: LHL = 2(2) + 3 = 7, RHL = 2(2) – 3 = 1. LHL ≠ RHL, so x = 2 is the only point of discontinuity.
f(x) = |x| + 3, if x ≤ –3; –2x, if –3 < x < 3; 6x + 2, if x ≥ 3
Answer
Each piece is continuous on its open interval. At x = –3: LHL = |–3| + 3 = 6, RHL = –2(–3) = 6, f(–3) = 6
⇒ continuous.
At x = 3: LHL = –2(3) = –6, RHL = 6(3) + 2 = 20
⇒ discontinuous. The only point of discontinuity is x = 3.
f(x) = |x|/x, if x ≠ 0; 0, if x = 0
Answer
f(x) = –1 for x < 0 and 1 for x > 0, so f is continuous for x ≠ 0. At x = 0: LHL = –1, RHL = 1
⇒ not equal. x = 0 is the only point of discontinuity.
f(x) = x/|x|, if x < 0; –1, if x ≥ 0
Answer
For x < 0, |x| = –x, so x/|x| = –1. Hence f(x) = –1 for all real x, a constant function. It is continuous everywhere; there is no point of discontinuity.
f(x) = x + 1, if x ≥ 1; x² + 1, if x < 1
Answer
Both pieces are polynomials. At x = 1: LHL = 1 + 1 = 2, RHL = 1 + 1 = 2, f(1) = 2
⇒ continuous. No point of discontinuity.
f(x) = x³ – 3, if x ≤ 2; x² + 1, if x > 2
Answer
At x = 2: LHL = 8 – 3 = 5, RHL = 4 + 1 = 5, f(2) = 5
⇒ continuous. Elsewhere f is a polynomial. No point of discontinuity.
f(x) = x¹⁰ – 1, if x ≤ 1; x², if x > 1
Answer
At x = 1: LHL = 1 – 1 = 0, RHL = 1
⇒ not equal. f is discontinuous only at x = 1.
Is the function defined by f(x) = x + 5, if x ≤ 1; x – 5, if x > 1 a continuous function?
Answer
At x = 1: LHL = 6, RHL = 1 – 5 = –4
⇒ LHL ≠ RHL. f is discontinuous at x = 1 (continuous elsewhere), so f is not a continuous function.
Discuss the continuity of the function f, where f is defined by (Exercises 14 to 16):
f(x) = 3, if 0 ≤ x ≤ 1; 4, if 1 < x < 3; 5, if 3 ≤ x ≤ 10
Answer
f is constant on each piece, so continuous inside each interval. At x = 1: LHL = 3, RHL = 4
⇒ discontinuous. At x = 3: LHL = 4, RHL = 5
⇒ discontinuous. So f is continuous on [0, 10] except at x = 1 and x = 3.
f(x) = 2x, if x < 0; 0, if 0 ≤ x ≤ 1; 4x, if x > 1
Answer
At x = 0: LHL = 0, RHL = 0, f(0) = 0
⇒ continuous. At x = 1: LHL = 0, RHL = 4
⇒ discontinuous. f is continuous everywhere except at x = 1.
f(x) = –2, if x ≤ –1; 2x, if –1 < x ≤ 1; 2, if x > 1
Answer
At x = –1: LHL = –2, RHL = 2(–1) = –2, f(–1) = –2
⇒ continuous. At x = 1: LHL = 2, RHL = 2, f(1) = 2
⇒ continuous. Each piece is continuous, so f is continuous at every real number.
Find the relationship between a and b so that the function f defined by f(x) = ax + 1, if x ≤ 3; bx + 3, if x > 3 is continuous at x = 3.
Answer
LHL = f(3) = 3a + 1, RHL = 3b + 3. For continuity: 3a + 1 = 3b + 3
⇒ 3a – 3b = 2
⇒ a = b + 2/3.
For what value of λ is the function defined by f(x) = λ(x² – 2x), if x ≤ 0; 4x + 1, if x > 0 continuous at x = 0? What about continuity at x = 1?
Answer
At x = 0: LHL = f(0) = λ(0) = 0, RHL = 4(0) + 1 = 1. Since 0 ≠ 1 whatever λ is, f is not continuous at x = 0 for any value of λ.
At x = 1: f(x) = 4x + 1 near 1, so lim = 5 = f(1). f is continuous at x = 1 for every λ.
Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Answer
Let n be an integer. For x slightly less than n, [x] = n – 1, so LHL = n – (n – 1) = 1. For x slightly more than n, [x] = n, so RHL = n – n = 0.
LHL ≠ RHL, so g is discontinuous at every integer n.
Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?
Answer
x², sin x and the constant 5 are continuous everywhere, so f is continuous. Check: lim(x
⇒ π) f(x) = π² – 0 + 5 = f(π). Yes, f is continuous at x = π.
Discuss the continuity of the following functions: (a) f(x) = sin x + cos x (b) f(x) = sin x – cos x (c) f(x) = sin x · cos x
Answer
sin x and cos x are continuous at every real number. The sum, difference and product of continuous functions are continuous, so all three functions are continuous everywhere.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer
cos x: lim(h
⇒ 0) cos(c + h) = cos c cos 0 – sin c sin 0 = cos c, so cos is continuous everywhere.
cosec x = 1/sin x: a quotient of continuous functions, continuous wherever sin x ≠ 0, i.e. for x ≠ nπ (n ∈ Z). It is continuous on its domain.
sec x = 1/cos x: continuous for all x ≠ (2n + 1)π/2.
cot x = cos x/sin x: continuous for all x ≠ nπ. So each function is continuous on its domain.
Find all points of discontinuity of f, where f(x) = (sin x)/x, if x < 0; x + 1, if x ≥ 0
Answer
For x < 0 and x > 0, f is continuous. At x = 0: LHL = lim(x
⇒ 0) sin x/x = 1, RHL = 0 + 1 = 1, f(0) = 1
⇒ continuous. f has no point of discontinuity.
Determine if f defined by f(x) = x² sin(1/x), if x ≠ 0; 0, if x = 0 is a continuous function?
Answer
For x ≠ 0, f is a product of continuous functions, so continuous. At x = 0: –1 ≤ sin(1/x) ≤ 1
⇒ –x² ≤ x² sin(1/x) ≤ x². As x
⇒ 0 both bounds
⇒ 0, so lim f(x) = 0 = f(0). f is continuous everywhere.
Examine the continuity of f, where f is defined by f(x) = sin x – cos x, if x ≠ 0; –1, if x = 0
Answer
For x ≠ 0, f is continuous. At x = 0: lim(x
⇒ 0) (sin x – cos x) = 0 – 1 = –1 = f(0). So f is continuous everywhere.
Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.
f(x) = (k cos x)/(π – 2x), if x ≠ π/2; 3, if x = π/2; at x = π/2
Answer
Put x = π/2 + h, h
⇒ 0: cos x = –sin h and π – 2x = –2h.
lim = lim(h
⇒ 0) k(–sin h)/(–2h) = (k/2) × 1 = k/2. Continuity needs k/2 = 3
⇒ k = 6.
f(x) = kx², if x ≤ 2; 3, if x > 2; at x = 2
Answer
LHL = f(2) = 4k, RHL = 3. 4k = 3
⇒ k = 3/4.
f(x) = kx + 1, if x ≤ π; cos x, if x > π; at x = π
Answer
LHL = f(π) = kπ + 1, RHL = cos π = –1. kπ + 1 = –1
⇒ k = –2/π.
f(x) = kx + 1, if x ≤ 5; 3x – 5, if x > 5; at x = 5
Answer
LHL = f(5) = 5k + 1, RHL = 15 – 5 = 10. 5k + 1 = 10
⇒ k = 9/5.
Find the values of a and b such that the function defined by f(x) = 5, if x ≤ 2; ax + b, if 2 < x < 10; 21, if x ≥ 10 is a continuous function.
Answer
At x = 2: 5 = 2a + b. At x = 10: 10a + b = 21. Subtracting: 8a = 16
⇒ a = 2
⇒ b = 1.
Show that the function defined by f(x) = cos(x²) is a continuous function.
Answer
f = g ∘ h with h(x) = x² and g(x) = cos x, both continuous everywhere. A composition of continuous functions is continuous, so cos(x²) is continuous.
Show that the function defined by f(x) = |cos x| is a continuous function.
Answer
f = g ∘ h with h(x) = cos x and g(x) = |x|, both continuous everywhere. Hence f = |cos x| is continuous.
Examine that sin |x| is a continuous function.
Answer
sin|x| = g ∘ h with h(x) = |x| and g(x) = sin x, both continuous. So sin|x| is continuous everywhere.
Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.
Answer
g(x) = |x| and h(x) = |x + 1| are continuous everywhere, so their difference is continuous everywhere.
(Check at the critical points: f(x) = 1 for x < –1, –2x – 1 for –1 ≤ x < 0, –1 for x ≥ 0; at x = –1 both sides give 1 and at x = 0 both give –1.) There is no point of discontinuity.
Exercise 5.2
Differentiate the functions with respect to x in Exercises 1 to 8.
sin(x² + 5)
Answer
By the chain rule: d/dx sin(x² + 5) = cos(x² + 5) × 2x = 2x cos(x² + 5).
cos(sin x)
Answer
–sin(sin x) × cos x = –cos x sin(sin x).
sin(ax + b)
Answer
cos(ax + b) × a = a cos(ax + b).
sec(tan(√x))
Answer
sec(tan √x) tan(tan √x) × sec²(√x) × 1/(2√x) = sec(tan √x) tan(tan √x) sec²√x / (2√x).
sin(ax + b)/cos(cx + d)
Answer
Quotient rule: [a cos(ax + b) cos(cx + d) + c sin(ax + b) sin(cx + d)] / cos²(cx + d)
= a cos(ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d).
cos x³ · sin²(x⁵)
Answer
Product rule: d/dx(cos x³) = –3x² sin x³ and d/dx(sin² x⁵) = 2 sin x⁵ cos x⁵ × 5x⁴ = 10x⁴ sin x⁵ cos x⁵.
So derivative = 10x⁴ sin x⁵ cos x⁵ cos x³ – 3x² sin x³ sin² x⁵.
2√(cot(x²))
Answer
2 × 1/(2√(cot x²)) × (–cosec² x²) × 2x = –2x cosec²(x²)/√(cot x²).
Simplifying, this equals –2√2 x / (sin x² √(sin 2x²)).
cos(√x)
Answer
–sin √x × 1/(2√x) = –sin √x/(2√x).
Prove that the function f given by f(x) = |x – 1|, x ∈ R is not differentiable at x = 1.
Answer
LHD at 1 = lim(h
⇒ 0⁻) [f(1 + h) – f(1)]/h = lim |h|/h = lim (–h)/h = –1.
RHD at 1 = lim(h
⇒ 0⁺) |h|/h = h/h = 1. LHD ≠ RHD, so f is not differentiable at x = 1.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
Answer
At x = 1: LHD = lim(h
⇒ 0⁻) ([1 + h] – [1])/h = lim (0 – 1)/h = lim (–1/h), which does not exist (
⇒ +∞). RHD = lim(h
⇒ 0⁺) (1 – 1)/h = 0. So f is not differentiable at x = 1.
At x = 2: LHD = lim(h
⇒ 0⁻) (1 – 2)/h does not exist, RHD = lim(h
⇒ 0⁺) (2 – 2)/h = 0. So f is not differentiable at x = 2.
Exercise 5.3
Find dy/dx in the following
2x + 3y = sin x
Answer
2 + 3y′ = cos x → y′ = (cos x – 2)/3.
2x + 3y = sin y
Answer
2 + 3y′ = cos y · y′
⇒ y′(cos y – 3) = 2
⇒ y′ = 2/(cos y – 3).
ax + by² = cos y
Answer
a + 2by y′ = –sin y · y′
⇒ y′(2by + sin y) = –a
⇒ y′ = –a/(2by + sin y).
xy + y² = tan x + y
Answer
y + xy′ + 2y y′ = sec²x + y′
⇒ y′(x + 2y – 1) = sec²x – y
⇒ y′ = (sec²x – y)/(x + 2y – 1).
x² + xy + y² = 100
Answer
2x + y + xy′ + 2y y′ = 0
⇒ y′(x + 2y) = –(2x + y)
⇒ y′ = –(2x + y)/(x + 2y).
x³ + x²y + xy² + y³ = 81
Answer
3x² + 2xy + x²y′ + y² + 2xy y′ + 3y² y′ = 0
⇒ y′(x² + 2xy + 3y²) = –(3x² + 2xy + y²)
⇒ y′ = –(3x² + 2xy + y²)/(x² + 2xy + 3y²).
sin² y + cos xy = k
Answer
2 sin y cos y · y′ – sin xy (y + xy′) = 0
⇒ y′(sin 2y – x sin xy) = y sin xy
⇒ y′ = y sin xy/(sin 2y – x sin xy).
sin²x + cos²y = 1
Answer
2 sin x cos x – 2 cos y sin y · y′ = 0
⇒ sin 2x = sin 2y · y′
⇒ y′ = sin 2x/sin 2y.
y = sin⁻¹(2x/(1 + x²))
Answer
Put x = tan θ: 2x/(1 + x²) = sin 2θ, so y = sin⁻¹(sin 2θ) = 2θ = 2 tan⁻¹x (for –1 ≤ x ≤ 1).
y′ = 2/(1 + x²).
y = tan⁻¹((3x – x³)/(1 – 3x²)), –1/√3 < x < 1/√3
Answer
Put x = tan θ, –π/6 < θ < π/6: (3 tan θ – tan³θ)/(1 – 3tan²θ) = tan 3θ, so y = 3θ = 3 tan⁻¹x.
y′ = 3/(1 + x²).
y = cos⁻¹((1 – x²)/(1 + x²)), 0 < x < 1
Answer
Put x = tan θ, 0 < θ < π/4: (1 – tan²θ)/(1 + tan²θ) = cos 2θ, so y = 2θ = 2 tan⁻¹x
⇒ y′ = 2/(1 + x²).
y = sin⁻¹((1 – x²)/(1 + x²)), 0 < x < 1
Answer
Put x = tan θ: sin y = cos 2θ = sin(π/2 – 2θ), with π/2 – 2θ in (0, π/2). So y = π/2 – 2 tan⁻¹x
⇒ y′ = –2/(1 + x²).
y = cos⁻¹(2x/(1 + x²)), –1 < x < 1
Answer
Put x = tan θ, –π/4 < θ < π/4: cos y = sin 2θ = cos(π/2 – 2θ), with π/2 – 2θ in (0, π). So y = π/2 – 2 tan⁻¹x
⇒ y′ = –2/(1 + x²).
y = sin⁻¹(2x√(1 – x²)), –1/√2 < x < 1/√2
Answer
Put x = sin θ, –π/4 < θ < π/4: 2x√(1 – x²) = 2 sin θ cos θ = sin 2θ, so y = 2θ = 2 sin⁻¹x
⇒ y′ = 2/√(1 – x²).
y = sec⁻¹(1/(2x² – 1)), 0 < x < 1/√2
Answer
sec y = 1/(2x² – 1)
⇒ cos y = 2x² – 1. Put x = cos θ, π/4 < θ < π/2: cos y = cos 2θ with 2θ in (π/2, π), so y = 2θ = 2 cos⁻¹x.
y′ = –2/√(1 – x²).
Exercise 5.4
Differentiate the following w.r.t. x
eˣ/sin x
Answer
(eˣ sin x – eˣ cos x)/sin²x = eˣ(sin x – cos x)/sin²x, x ≠ nπ.
e^(sin⁻¹x)
Answer
e^(sin⁻¹x) × 1/√(1 – x²) = e^(sin⁻¹x)/√(1 – x²), x ∈ (–1, 1).
e^(x³)
Answer
e^(x³) × 3x² = 3x² e^(x³).
sin(tan⁻¹ e⁻ˣ)
Answer
cos(tan⁻¹e⁻ˣ) × 1/(1 + e⁻²ˣ) × (–e⁻ˣ) = –e⁻ˣ cos(tan⁻¹e⁻ˣ)/(1 + e⁻²ˣ).
log(cos eˣ)
Answer
(1/cos eˣ) × (–sin eˣ) × eˣ = –eˣ tan eˣ, eˣ ≠ (2n + 1)π/2.
eˣ + e^(x²) + … + e^(x⁵)
Answer
eˣ + 2x e^(x²) + 3x² e^(x³) + 4x³ e^(x⁴) + 5x⁴ e^(x⁵).
√(e^(√x)), x > 0
Answer
y = e^(√x/2)
⇒ y′ = e^(√x/2) × 1/(4√x) = √(e^(√x))/(4√x).
log(log x), x > 1
Answer
(1/log x) × (1/x) = 1/(x log x).
cos x/log x, x > 0
Answer
[–sin x log x – cos x × (1/x)]/(log x)² = –(x log x sin x + cos x)/(x (log x)²).
cos(log x + eˣ), x > 0
Answer
–sin(log x + eˣ) × (1/x + eˣ) = –(1/x + eˣ) sin(log x + eˣ).
Exercise 5.5
Differentiate the functions given in Exercises 1 to 11 w.r.t. x.
cos x · cos 2x · cos 3x
Answer
log y = log cos x + log cos 2x + log cos 3x
⇒ y′/y = –tan x – 2 tan 2x – 3 tan 3x.
y′ = –cos x cos 2x cos 3x (tan x + 2 tan 2x + 3 tan 3x).
√[(x – 1)(x – 2)/((x – 3)(x – 4)(x – 5))]
Answer
log y = (1/2)[log(x – 1) + log(x – 2) – log(x – 3) – log(x – 4) – log(x – 5)].
y′ = (1/2) √[(x – 1)(x – 2)/((x – 3)(x – 4)(x – 5))] × [1/(x – 1) + 1/(x – 2) – 1/(x – 3) – 1/(x – 4) – 1/(x – 5)].
(log x)^(cos x)
Answer
log y = cos x · log(log x)
⇒ y′/y = –sin x log(log x) + cos x/(x log x).
y′ = (log x)^(cos x) [cos x/(x log x) – sin x log(log x)].
xˣ – 2^(sin x)
Answer
u = xˣ: log u = x log x
⇒ u′ = xˣ(1 + log x). v = 2^(sin x): v′ = 2^(sin x) log 2 × cos x.
y′ = xˣ(1 + log x) – 2^(sin x) cos x log 2.
(x + 3)² · (x + 4)³ · (x + 5)⁴
Answer
log y = 2 log(x + 3) + 3 log(x + 4) + 4 log(x + 5)
⇒ y′ = y[2/(x + 3) + 3/(x + 4) + 4/(x + 5)].
Simplifying: y′ = (x + 3)(x + 4)²(x + 5)³ (9x² + 70x + 133).
(x + 1/x)ˣ + x^(1 + 1/x)
Answer
u = (x + 1/x)ˣ: log u = x log(x + 1/x)
⇒ u′/u = log(x + 1/x) + x × (1 – 1/x²)/(x + 1/x) = log(x + 1/x) + (x² – 1)/(x² + 1).
v = x^(1 + 1/x): log v = (1 + 1/x) log x
⇒ v′/v = –log x/x² + (1 + 1/x)/x = (x + 1 – log x)/x².
y′ = (x + 1/x)ˣ [(x² – 1)/(x² + 1) + log(x + 1/x)] + x^(1 + 1/x) (x + 1 – log x)/x².
(log x)ˣ + x^(log x)
Answer
u = (log x)ˣ: log u = x log(log x)
⇒ u′ = (log x)ˣ [log(log x) + 1/log x].
v = x^(log x): log v = (log x)²
⇒ v′ = x^(log x) × 2 log x/x.
y′ = (log x)ˣ [log(log x) + 1/log x] + 2x^(log x – 1) log x.
(sin x)ˣ + sin⁻¹√x
Answer
u = (sin x)ˣ: log u = x log sin x
⇒ u′ = (sin x)ˣ (log sin x + x cot x).
d/dx sin⁻¹√x = 1/√(1 – x) × 1/(2√x) = 1/(2√(x – x²)).
y′ = (sin x)ˣ (x cot x + log sin x) + 1/(2√(x – x²)).
x^(sin x) + (sin x)^(cos x)
Answer
u = x^(sin x): log u = sin x log x
⇒ u′ = x^(sin x) (cos x log x + sin x/x).
v = (sin x)^(cos x): log v = cos x log sin x
⇒ v′ = (sin x)^(cos x) (cos x cot x – sin x log sin x).
y′ = x^(sin x) (cos x log x + sin x/x) + (sin x)^(cos x) (cos x cot x – sin x log sin x).
x^(x cos x) + (x² + 1)/(x² – 1)
Answer
u = x^(x cos x): log u = x cos x log x
⇒ u′/u = (cos x – x sin x) log x + cos x.
d/dx (x² + 1)/(x² – 1) = [2x(x² – 1) – 2x(x² + 1)]/(x² – 1)² = –4x/(x² – 1)².
y′ = x^(x cos x) [cos x (1 + log x) – x sin x log x] – 4x/(x² – 1)².
(x cos x)ˣ + (x sin x)^(1/x)
Answer
u = (x cos x)ˣ: log u = x[log x + log cos x]
⇒ u′/u = log(x cos x) + x(1/x – tan x) = 1 – x tan x + log(x cos x).
v = (x sin x)^(1/x): log v = (1/x) log(x sin x)
⇒ v′/v = –log(x sin x)/x² + (1/x)(1/x + cot x) = [1 + x cot x – log(x sin x)]/x².
y′ = (x cos x)ˣ [1 – x tan x + log(x cos x)] + (x sin x)^(1/x) [1 + x cot x – log(x sin x)]/x².
Find dy/dx of the functions given in Exercises 12 to 15.
xʸ + yˣ = 1
Answer
u = xʸ: log u = y log x
⇒ u′ = xʸ (y/x + y′ log x). v = yˣ: log v = x log y
⇒ v′ = yˣ (log y + x y′/y).
u′ + v′ = 0
⇒ y′(xʸ log x + x yˣ⁻¹) = –(y xʸ⁻¹ + yˣ log y)
⇒ y′ = –(y xʸ⁻¹ + yˣ log y)/(xʸ log x + x yˣ⁻¹).
yˣ = xʸ
Answer
x log y = y log x
⇒ log y + (x/y)y′ = y′ log x + y/x
⇒ y′(x/y – log x) = y/x – log y.
y′ = y(y – x log y)/(x(x – y log x)).
(cos x)ʸ = (cos y)ˣ
Answer
y log cos x = x log cos y
⇒ y′ log cos x – y tan x = log cos y – x tan y · y′.
y′(log cos x + x tan y) = log cos y + y tan x
⇒ y′ = (y tan x + log cos y)/(x tan y + log cos x).
xy = e^(x – y)
Answer
log x + log y = x – y
⇒ 1/x + y′/y = 1 – y′
⇒ y′(1/y + 1) = (x – 1)/x
⇒ y′ = y(x – 1)/(x(y + 1)).
Find the derivative of the function given by f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) and hence find f′(1).
Answer
log f = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸).
f′(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) [1/(1 + x) + 2x/(1 + x²) + 4x³/(1 + x⁴) + 8x⁷/(1 + x⁸)].
f(1) = 2 × 2 × 2 × 2 = 16
⇒ f′(1) = 16 (1/2 + 1 + 2 + 4) = 16 × 15/2 = 120.
Differentiate (x² – 5x + 8)(x³ + 7x + 9) in three ways mentioned below: (i) by using product rule (ii) by expanding the product to obtain a single polynomial (iii) by logarithmic differentiation. Do they all give the same answer?
Answer
(i) (2x – 5)(x³ + 7x + 9) + (x² – 5x + 8)(3x² + 7) = (2x⁴ – 5x³ + 14x² – 17x – 45) + (3x⁴ – 15x³ + 31x² – 35x + 56) = 5x⁴ – 20x³ + 45x² – 52x + 11.
(ii) Product = x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72
⇒ derivative = 5x⁴ – 20x³ + 45x² – 52x + 11.
(iii) log y = log(x² – 5x + 8) + log(x³ + 7x + 9)
⇒ y′ = y[(2x – 5)/(x² – 5x + 8) + (3x² + 7)/(x³ + 7x + 9)] = (2x – 5)(x³ + 7x + 9) + (3x² + 7)(x² – 5x + 8) = 5x⁴ – 20x³ + 45x² – 52x + 11.
Yes, all three methods give the same answer.
If u, v and w are functions of x, then show that d/dx(u · v · w) = (du/dx) v · w + u · (dv/dx) · w + u · v (dw/dx) in two ways – first by repeated application of product rule, second by logarithmic differentiation.
Answer
Method 1: d/dx[(uv)w] = (uv)′w + (uv)w′ = (u′v + uv′)w + uvw′ = u′vw + uv′w + uvw′.
Method 2: Let y = uvw. log y = log u + log v + log w
⇒ y′/y = u′/u + v′/v + w′/w
⇒ y′ = uvw(u′/u + v′/v + w′/w) = u′vw + uv′w + uvw′. Hence proved.
Exercise 5.6
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx.
x = 2at², y = at⁴
Answer
dx/dt = 4at, dy/dt = 4at³
⇒ dy/dx = 4at³/(4at) = t².
x = a cos θ, y = b cos θ
Answer
dx/dθ = –a sin θ, dy/dθ = –b sin θ
⇒ dy/dx = b/a.
x = sin t, y = cos 2t
Answer
dx/dt = cos t, dy/dt = –2 sin 2t
⇒ dy/dx = –2 sin 2t/cos t = –4 sin t cos t/cos t = –4 sin t.
x = 4t, y = 4/t
Answer
dx/dt = 4, dy/dt = –4/t²
⇒ dy/dx = –1/t².
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Answer
dx/dθ = –sin θ + 2 sin 2θ, dy/dθ = cos θ – 2 cos 2θ
⇒ dy/dx = (cos θ – 2 cos 2θ)/(2 sin 2θ – sin θ).
x = a(θ – sin θ), y = a(1 + cos θ)
Answer
dx/dθ = a(1 – cos θ), dy/dθ = –a sin θ
⇒ dy/dx = –sin θ/(1 – cos θ) = –2 sin(θ/2) cos(θ/2)/(2 sin²(θ/2)) = –cot(θ/2).
x = sin³t/√(cos 2t), y = cos³t/√(cos 2t)
Answer
Quotient rule (using d/dt √(cos 2t) = –sin 2t/√(cos 2t)): dx/dt = [3 sin²t cos t cos 2t + sin³t sin 2t]/(cos 2t)^(3/2).
Numerator = sin²t cos t (3 cos 2t + 2 sin²t) = sin²t cos t (3 – 4 sin²t) = sin t cos t sin 3t. So dx/dt = sin t cos t sin 3t/(cos 2t)^(3/2).
dy/dt = [–3 cos²t sin t cos 2t + cos³t sin 2t]/(cos 2t)^(3/2). Numerator = sin t cos²t (2cos²t – 3 cos 2t) = sin t cos²t (3 – 4cos²t) = –sin t cos t cos 3t. So dy/dt = –sin t cos t cos 3t/(cos 2t)^(3/2).
dy/dx = (dy/dt)/(dx/dt) = –cos 3t/sin 3t = –cot 3t.
x = a(cos t + log tan(t/2)), y = a sin t
Answer
d/dt log tan(t/2) = sec²(t/2)/(2 tan(t/2)) = 1/(2 sin(t/2) cos(t/2)) = 1/sin t.
dx/dt = a(–sin t + 1/sin t) = a cos²t/sin t, dy/dt = a cos t
⇒ dy/dx = a cos t × sin t/(a cos²t) = tan t.
x = a sec θ, y = b tan θ
Answer
dx/dθ = a sec θ tan θ, dy/dθ = b sec²θ
⇒ dy/dx = b sec θ/(a tan θ) = (b/a) cosec θ.
x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ)
Answer
dx/dθ = a(–sin θ + sin θ + θ cos θ) = aθ cos θ, dy/dθ = a(cos θ – cos θ + θ sin θ) = aθ sin θ
⇒ dy/dx = tan θ.
If x = √(a^(sin⁻¹t)), y = √(a^(cos⁻¹t)), show that dy/dx = –y/x
Answer
log x = (1/2) sin⁻¹t log a
⇒ dx/dt = x log a/(2√(1 – t²)). log y = (1/2) cos⁻¹t log a
⇒ dy/dt = –y log a/(2√(1 – t²)).
dy/dx = (dy/dt)/(dx/dt) = –y/x. Hence shown.
Exercise 5.7
Find the second order derivatives of the functions given in Exercises 1 to 10.
x² + 3x + 2
Answer
y′ = 2x + 3 → y″ = 2.
x²⁰
Answer
y′ = 20x¹⁹ → y″ = 380x¹⁸.
x · cos x
Answer
y′ = cos x – x sin x
⇒ y″ = –sin x – sin x – x cos x = –(x cos x + 2 sin x).
log x
Answer
y′ = 1/x → y″ = –1/x².
x³ log x
Answer
y′ = 3x² log x + x²
⇒ y″ = 6x log x + 3x + 2x = x(5 + 6 log x).
eˣ sin 5x
Answer
y′ = eˣ(sin 5x + 5 cos 5x)
⇒ y″ = eˣ(sin 5x + 5 cos 5x) + eˣ(5 cos 5x – 25 sin 5x) = 2eˣ(5 cos 5x – 12 sin 5x).
e⁶ˣ cos 3x
Answer
y′ = e⁶ˣ(6 cos 3x – 3 sin 3x)
⇒ y″ = 6e⁶ˣ(6 cos 3x – 3 sin 3x) + e⁶ˣ(–18 sin 3x – 9 cos 3x) = 9e⁶ˣ(3 cos 3x – 4 sin 3x).
tan⁻¹x
Answer
y′ = 1/(1 + x²) → y″ = –2x/(1 + x²)².
log(log x)
Answer
y′ = 1/(x log x) = (x log x)⁻¹
⇒ y″ = –(log x + 1)/(x log x)².
sin(log x)
Answer
y′ = cos(log x)/x
⇒ y″ = [–sin(log x) × (1/x) × x – cos(log x)]/x² = –[sin(log x) + cos(log x)]/x².
If y = 5 cos x – 3 sin x, prove that d²y/dx² + y = 0
Answer
y′ = –5 sin x – 3 cos x
⇒ y″ = –5 cos x + 3 sin x = –y. So y″ + y = 0.
If y = cos⁻¹x, Find d²y/dx² in terms of y alone.
Answer
y′ = –(1 – x²)^(–1/2)
⇒ y″ = –(1/2)(1 – x²)^(–3/2)(2x) = –x/(1 – x²)^(3/2).
With x = cos y and 1 – x² = sin²y: y″ = –cos y/sin³y = –cot y cosec²y.
If y = 3 cos(log x) + 4 sin(log x), show that x²y₂ + xy₁ + y = 0
Answer
y₁ = [–3 sin(log x) + 4 cos(log x)]/x
⇒ xy₁ = –3 sin(log x) + 4 cos(log x).
Differentiate: xy₂ + y₁ = [–3 cos(log x) – 4 sin(log x)]/x = –y/x
⇒ x²y₂ + xy₁ + y = 0.
If y = Ae^(mx) + Be^(nx), show that d²y/dx² – (m + n) dy/dx + mny = 0
Answer
y′ = Ame^(mx) + Bne^(nx), y″ = Am²e^(mx) + Bn²e^(nx).
y″ – (m + n)y′ + mny = Ae^(mx)(m² – m² – mn + mn) + Be^(nx)(n² – mn – n² + mn) = 0.
If y = 500e^(7x) + 600e^(–7x), show that d²y/dx² = 49y
Answer
y′ = 3500e^(7x) – 4200e^(–7x)
⇒ y″ = 24500e^(7x) + 29400e^(–7x) = 49(500e^(7x) + 600e^(–7x)) = 49y.
If eʸ(x + 1) = 1, show that d²y/dx² = (dy/dx)²
Answer
eʸ = 1/(x + 1)
⇒ y = –log(x + 1)
⇒ y′ = –1/(x + 1)
⇒ y″ = 1/(x + 1)² = (y′)².
If y = (tan⁻¹x)², show that (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2
Answer
y₁ = 2 tan⁻¹x/(1 + x²)
⇒ (1 + x²)y₁ = 2 tan⁻¹x.
Differentiate: (1 + x²)y₂ + 2xy₁ = 2/(1 + x²). Multiply by (1 + x²): (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2.
Miscellaneous Exercise on Chapter 5
Differentiate w.r.t. x the function in Exercises 1 to 11.
(3x² – 9x + 5)⁹
Answer
9(3x² – 9x + 5)⁸(6x – 9) = 27(3x² – 9x + 5)⁸(2x – 3).
sin³x + cos⁶x
Answer
3 sin²x cos x + 6 cos⁵x(–sin x) = 3 sin x cos x (sin x – 2 cos⁴x).
(5x)^(3 cos 2x)
Answer
log y = 3 cos 2x log 5x
⇒ y′/y = –6 sin 2x log 5x + 3 cos 2x/x.
y′ = (5x)^(3 cos 2x) [3 cos 2x/x – 6 sin 2x log 5x].
sin⁻¹(x√x), 0 ≤ x ≤ 1
Answer
y = sin⁻¹(x^(3/2))
⇒ y′ = (3/2)x^(1/2)/√(1 – x³) = (3/2)√(x/(1 – x³)).
cos⁻¹(x/2)/√(2x + 7), –2 < x < 2
Answer
d/dx cos⁻¹(x/2) = –1/√(4 – x²). Quotient rule:
y′ = [–√(2x + 7)/√(4 – x²) – cos⁻¹(x/2)/√(2x + 7)]/(2x + 7) = –[1/(√(4 – x²) √(2x + 7)) + cos⁻¹(x/2)/(2x + 7)^(3/2)].
cot⁻¹[(√(1 + sin x) + √(1 – sin x))/(√(1 + sin x) – √(1 – sin x))], 0 < x < π/2
Answer
For 0 < x < π/2: √(1 + sin x) = cos(x/2) + sin(x/2) and √(1 – sin x) = cos(x/2) – sin(x/2).
Fraction = 2cos(x/2)/(2 sin(x/2)) = cot(x/2)
⇒ y = cot⁻¹(cot(x/2)) = x/2
⇒ y′ = 1/2.
(log x)^(log x), x > 1
Answer
log y = log x · log(log x)
⇒ y′/y = (1/x) log(log x) + log x × 1/(x log x) = [1 + log(log x)]/x.
y′ = (log x)^(log x) [1 + log(log x)]/x.
cos(a cos x + b sin x), for some constant a and b.
Answer
–sin(a cos x + b sin x) × (–a sin x + b cos x) = (a sin x – b cos x) sin(a cos x + b sin x).
(sin x – cos x)^(sin x – cos x), π/4 < x < 3π/4
Answer
log y = (sin x – cos x) log(sin x – cos x)
⇒ y′/y = (cos x + sin x) log(sin x – cos x) + (cos x + sin x).
y′ = (sin x – cos x)^(sin x – cos x) (cos x + sin x)[1 + log(sin x – cos x)].
xˣ + xᵃ + aˣ + aᵃ, for some fixed a > 0 and x > 0
Answer
xˣ(1 + log x) + a xᵃ⁻¹ + aˣ log a + 0.
x^(x² – 3) + (x – 3)^(x²), for x > 3
Answer
u = x^(x² – 3): log u = (x² – 3) log x
⇒ u′ = x^(x² – 3) [(x² – 3)/x + 2x log x].
v = (x – 3)^(x²): log v = x² log(x – 3)
⇒ v′ = (x – 3)^(x²) [x²/(x – 3) + 2x log(x – 3)].
y′ = x^(x² – 3) [(x² – 3)/x + 2x log x] + (x – 3)^(x²) [x²/(x – 3) + 2x log(x – 3)].
Find dy/dx, if y = 12(1 – cos t), x = 10(t – sin t), –π/2 < t < π/2
Answer
dy/dt = 12 sin t, dx/dt = 10(1 – cos t)
⇒ dy/dx = 12 sin t/(10(1 – cos t)) = (6/5) × 2 sin(t/2) cos(t/2)/(2 sin²(t/2)) = (6/5) cot(t/2).
Find dy/dx, if y = sin⁻¹x + sin⁻¹√(1 – x²), 0 < x < 1
Answer
For 0 < x < 1, sin⁻¹√(1 – x²) = cos⁻¹x. So y = sin⁻¹x + cos⁻¹x = π/2
⇒ dy/dx = 0.
(Directly: 1/√(1 – x²) + [1/√(1 – (1 – x²))] × (–x/√(1 – x²)) = 1/√(1 – x²) – 1/√(1 – x²) = 0.)
If x√(1 + y) + y√(1 + x) = 0, for –1 < x < 1, prove that dy/dx = –1/(1 + x)²
Answer
x√(1 + y) = –y√(1 + x)
⇒ x²(1 + y) = y²(1 + x)
⇒ x² – y² + xy(x – y) = 0
⇒ (x – y)(x + y + xy) = 0.
x = y does not satisfy the given equation (it gives 2x√(1 + x) = 0 only at x = 0), so x + y + xy = 0
⇒ y = –x/(1 + x).
dy/dx = –[(1 + x) – x]/(1 + x)² = –1/(1 + x)². Hence proved.
If (x – a)² + (y – b)² = c², for some c > 0, prove that [1 + (dy/dx)²]^(3/2)/(d²y/dx²) is a constant independent of a and b.
Answer
2(x – a) + 2(y – b)y′ = 0
⇒ y′ = –(x – a)/(y – b).
y″ = –[(y – b) – (x – a)y′]/(y – b)² = –[(y – b)² + (x – a)²]/(y – b)³ = –c²/(y – b)³.
1 + y′² = [(y – b)² + (x – a)²]/(y – b)² = c²/(y – b)²
⇒ [1 + y′²]^(3/2) = c³/(y – b)³.
Ratio = [c³/(y – b)³] ÷ [–c²/(y – b)³] = –c, a constant independent of a and b.
If cos y = x cos(a + y), with cos a ≠ ±1, prove that dy/dx = cos²(a + y)/sin a.
Answer
x = cos y/cos(a + y)
⇒ dx/dy = [–sin y cos(a + y) + cos y sin(a + y)]/cos²(a + y) = sin(a + y – y)/cos²(a + y) = sin a/cos²(a + y).
dy/dx = cos²(a + y)/sin a. Hence proved.
If x = a(cos t + t sin t) and y = a(sin t – t cos t), find d²y/dx².
Answer
dx/dt = at cos t, dy/dt = at sin t
⇒ dy/dx = tan t.
d²y/dx² = d/dt(tan t) × dt/dx = sec²t/(at cos t) = sec³t/(at), 0 < t < π/2.
If f(x) = |x|³, show that f″(x) exists for all real x and find it.
Answer
For x ≥ 0, f(x) = x³
⇒ f′(x) = 3x², f″(x) = 6x. For x < 0, f(x) = –x³
⇒ f′(x) = –3x², f″(x) = –6x.
At x = 0: f′(0) = lim(h
⇒ 0) |h|³/h = 0 from both sides, so f′(x) = 3x|x| for all x. f″(0) = lim(h
⇒ 0) 3h|h|/h = lim 3|h| = 0 from both sides.
So f″(x) exists everywhere: f″(x) = 6x for x ≥ 0 and –6x for x < 0, i.e. f″(x) = 6|x|.
Using the fact that sin(A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Answer
Treat B as a constant and differentiate both sides with respect to A:
cos(A + B) × 1 = cos A cos B + (–sin A) sin B
⇒ cos(A + B) = cos A cos B – sin A sin B.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Answer
Yes. f(x) = |x| + |x – 1| is continuous everywhere (sum of continuous functions).
f(x) = 1 – 2x for x < 0, 1 for 0 ≤ x < 1 and 2x – 1 for x ≥ 1. At x = 0: LHD = –2, RHD = 0
⇒ not differentiable. At x = 1: LHD = 0, RHD = 2
⇒ not differentiable. At every other point f is linear, hence differentiable.
If y = determinant with rows (f(x), g(x), h(x)), (l, m, n), (a, b, c), prove that dy/dx = determinant with rows (f′(x), g′(x), h′(x)), (l, m, n), (a, b, c).
Answer
Expanding along R₁: y = f(x)(mc – nb) – g(x)(lc – na) + h(x)(lb – ma). The brackets are constants.
dy/dx = f′(x)(mc – nb) – g′(x)(lc – na) + h′(x)(lb – ma), which is exactly the expansion along R₁ of the determinant with rows (f′(x), g′(x), h′(x)), (l, m, n), (a, b, c). Hence proved.
If y = e^(a cos⁻¹x), –1 ≤ x ≤ 1, show that (1 – x²) d²y/dx² – x dy/dx – a²y = 0.
Answer
y₁ = e^(a cos⁻¹x) × (–a/√(1 – x²))
⇒ √(1 – x²) y₁ = –ay.
Differentiate: √(1 – x²) y₂ – x y₁/√(1 – x²) = –a y₁. Multiply by √(1 – x²): (1 – x²)y₂ – xy₁ = –a√(1 – x²) y₁ = –a(–ay) = a²y.
So (1 – x²)y₂ – xy₁ – a²y = 0. Hence shown.
Important Extra Questions
Find k if f(x) = (sin 3x)/x for x ≠ 0 and f(0) = k is continuous at x = 0.
Answer
lim(x
⇒ 0) sin 3x/x = 3 × lim sin 3x/(3x) = 3. So k = 3.
Is every continuous function differentiable? Give an example.
Answer
No. f(x) = |x| is continuous at x = 0, but LHD = –1 and RHD = 1, so it is not differentiable there. (Every differentiable function is continuous, but not conversely.)
Differentiate y = xˣ w.r.t. x.
Answer
log y = x log x
⇒ y′/y = log x + 1
⇒ y′ = xˣ(1 + log x).
Find dy/dx if x = at², y = 2at.
Answer
dx/dt = 2at, dy/dt = 2a
⇒ dy/dx = 2a/(2at) = 1/t.
If y = A sin x + B cos x, show that y″ + y = 0.
Answer
y′ = A cos x – B sin x
⇒ y″ = –A sin x – B cos x = –y
⇒ y″ + y = 0.