Miscellaneous Exercise answers: Limits and Derivatives

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(i) −x

1. By first principle, f′(x) = lim(h→0) [f(x + h) − f(x)]/h.
2. f(x + h) − f(x) = −(x + h) − (−x) = −h.
3. Divide by h to get −1, which does not depend on h at all, so the limit is −1.

Answer−1

(ii) (−x)⁻¹

1. (−x)⁻¹ = −1/x. By first principle, f′(x) = lim(h→0) [f(x + h) − f(x)]/h.
2. f(x + h) − f(x) = −1/(x + h) − (−1/x) = [−x + (x + h)]/[x(x + h)] = h/[x(x + h)].
3. Divide by h: 1/[x(x + h)]. As h→0 this goes to 1/x².

Answer1/x²

(iii) sin (x + 1)

1. By first principle, f′(x) = lim(h→0) [sin(x + 1 + h) − sin(x + 1)]/h.
2. Use sin A − sin B = 2 cos((A+B)/2) sin((A−B)/2) with A = x + 1 + h, B = x + 1.
3. This gives 2 cos(x + 1 + h/2) sin(h/2), so the quotient is cos(x + 1 + h/2) · [sin(h/2)/(h/2)].
4. As h→0, this goes to cos(x + 1) · 1 = cos(x + 1).

Answercos (x + 1)

(iv) cos (x − π/8)

1. By first principle, f′(x) = lim(h→0) [cos(x − π/8 + h) − cos(x − π/8)]/h.
2. Use cos A − cos B = −2 sin((A+B)/2) sin((A−B)/2) with A = x − π/8 + h, B = x − π/8.
3. This gives −2 sin(x − π/8 + h/2) sin(h/2), so the quotient is −sin(x − π/8 + h/2) · [sin(h/2)/(h/2)].
4. As h→0, this goes to −sin(x − π/8).

Answer−sin (x − π/8)

Watch this explained “A difference becomes a product”, 2:39 into Sine and tangent go back to the definition, because no rule so far reaches them

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1. The derivative of x is 1, and the derivative of the constant a is 0, so the total is 1.

Answer1

Watch this explained “Why the constant term goes”, 10:03 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. Differentiate: (px + q)(r/x + s).
2. Use the product rule: (uv)′ = u′v + uv′, with u = px + q and v = r/x + s.
3. u′ = p, and v′ is the derivative of r/x + s, which is −r/x².
4. So the derivative is p(r/x + s) + (px + q)(−r/x²).
5. Multiply out and collect: pr/x + ps − pr/x − qr/x² = ps − qr/x².

Answerps − qr/x²

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Use the product rule with u = ax + b and v = (cx + d)²: (uv)′ = u′v + uv′.
2. u′ = a. For v, use the product rule again on (cx + d)(cx + d): v′ = c(cx + d) + (cx + d)c = 2c(cx + d).
3. So the derivative is a(cx + d)² + (ax + b) · 2c(cx + d).
4. Take out the common factor (cx + d): (cx + d)[a(cx + d) + 2c(ax + b)].

Answer(cx + d)[a(cx + d) + 2c(ax + b)]

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Use the quotient rule: (u/v)′ = (u′v − uv′)/v², with u = ax + b, v = cx + d.
2. u′ = a and v′ = c.
3. u′v − uv′ = a(cx + d) − c(ax + b) = acx + ad − acx − bc = ad − bc.
4. So the derivative is (ad − bc)/(cx + d)².

Answer(ad − bc)/(cx + d)²

Watch this explained “A worked ratio, and where it stops”, 11:59 into Rules for differentiating a sum, a product and a quotient

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1. Multiply the top and bottom by x to clear the small fractions: (1 + 1/x)·x = x + 1, and (1 − 1/x)·x = x − 1.
2. So the function is the same as (x + 1)/(x − 1), wherever the original is defined.
3. Use the quotient rule with u = x + 1, v = x − 1: u′ = 1, v′ = 1.
4. u′v − uv′ = (x − 1) − (x + 1) = −2, so the derivative is −2/(x − 1)².

Answer−2/(x − 1)²

Watch this explained “A worked ratio, and where it stops”, 11:59 into Rules for differentiating a sum, a product and a quotient

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1. Use the quotient rule with u = 1, v = ax² + bx + c.
2. u′ = 0 and v′ = 2ax + b.
3. u′v − uv′ = 0 − (2ax + b) = −(2ax + b).
4. So the derivative is −(2ax + b)/(ax² + bx + c)².

Answer−(2ax + b)/(ax² + bx + c)²

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Use the quotient rule with u = ax + b, v = px² + qx + r.
2. u′ = a and v′ = 2px + q.
3. u′v − uv′ = a(px² + qx + r) − (ax + b)(2px + q).
4. Put this over v² = (px² + qx + r)².

Answer[a(px² + qx + r) − (ax + b)(2px + q)]/(px² + qx + r)²

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Use the quotient rule with u = px² + qx + r, v = ax + b.
2. u′ = 2px + q and v′ = a.
3. u′v − uv′ = (2px + q)(ax + b) − (px² + qx + r)·a.
4. Put this over v² = (ax + b)².

Answer[(2px + q)(ax + b) − a(px² + qx + r)]/(ax + b)²

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Differentiate: a/x⁴ − b/x² + cos x.
2. Write a/x⁴ as ax⁻⁴ and b/x² as bx⁻².
3. Differentiate each term: ax⁻⁴ gives −4ax⁻⁵, −bx⁻² gives 2bx⁻³, and cos x gives −sin x.
4. Add the three results.

Answer−4a/x⁵ + 2b/x³ − sin x

Watch this explained “Below zero, where neither proof reaches”, 7:50 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. Write 4√x as 4x^(1/2).
2. Use the power rule: the derivative of x^(1/2) is (1/2)x^(−1/2), so 4x^(1/2) gives 4·(1/2)x^(−1/2) = 2x^(−1/2).
3. The derivative of the constant −2 is 0.

Answer2/√x

Watch this explained “Below zero, where neither proof reaches”, 7:50 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. Start with a small power to see the pattern. (ax + b)² = (ax + b)(ax + b), so the product rule gives a(ax + b) + (ax + b)·a = 2a(ax + b).
2. Go up one power at a time. Write (ax + b)ⁿ = (ax + b)·(ax + b)ⁿ⁻¹. If the power below has derivative (n − 1)a(ax + b)ⁿ⁻², the product rule gives a(ax + b)ⁿ⁻¹ + (ax + b)·(n − 1)a(ax + b)ⁿ⁻² = a(ax + b)ⁿ⁻¹ + (n − 1)a(ax + b)ⁿ⁻¹ = na(ax + b)ⁿ⁻¹.
3. n may also be a negative integer, n = −k. Then (ax + b)ⁿ = 1/(ax + b)ᵏ, and the quotient rule gives [0·(ax + b)ᵏ − 1·ka(ax + b)ᵏ⁻¹]/(ax + b)²ᵏ = −ka(ax + b)⁻ᵏ⁻¹ = na(ax + b)ⁿ⁻¹. (For n = 0 the function is the constant 1, and both sides are 0.)
4. So in every case: bring the power n down, lower the power by one, and multiply by a, the derivative of ax + b.

Answerna(ax + b)ⁿ⁻¹

Watch this explained “Second proof: expand nothing”, 4:01 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. Use the product rule (uv)′ = u′v + uv′ with u = (ax + b)ⁿ and v = (cx + d)ᵐ.
2. From Question 12, u′ = na(ax + b)ⁿ⁻¹. In the same way, with c, d and m in place of a, b and n, v′ = mc(cx + d)ᵐ⁻¹.
3. So the derivative is u′v + uv′ = na(ax + b)ⁿ⁻¹(cx + d)ᵐ + mc(ax + b)ⁿ(cx + d)ᵐ⁻¹.

Answerna(ax + b)ⁿ⁻¹(cx + d)ᵐ + mc(ax + b)ⁿ(cx + d)ᵐ⁻¹

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. By first principle, f′(x) = lim(h→0) [sin(x + a + h) − sin(x + a)]/h.
2. Use sin A − sin B = 2 cos((A+B)/2) sin((A−B)/2) with A = x + a + h, B = x + a.
3. This gives 2 cos(x + a + h/2) sin(h/2), so the quotient is cos(x + a + h/2) · [sin(h/2)/(h/2)].
4. As h→0 this goes to cos(x + a).

Answercos (x + a)

Watch this explained “A difference becomes a product”, 2:39 into Sine and tangent go back to the definition, because no rule so far reaches them

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1. Write y = cosec x × cot x. This is a product of two functions, so use the product rule: (uv)′ = u′v + uv′.
2. Here u = cosec x, so u′ = −cosec x cot x. And v = cot x, so v′ = −cosec²x.
3. So y′ = (−cosec x cot x)(cot x) + (cosec x)(−cosec²x) = −cosec x cot²x − cosec³x.
4. Take −cosec x common: y′ = −cosec x (cot²x + cosec²x).

Answer−cosec x (cosec²x + cot²x)

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = cos x / (1 + sin x). This is a quotient, so use the quotient rule: (u/v)′ = (u′v − uv′) / v².
2. Here u = cos x, so u′ = −sin x. And v = 1 + sin x, so v′ = cos x.
3. Numerator = (−sin x)(1 + sin x) − (cos x)(cos x) = −sin x − sin²x − cos²x.
4. Since sin²x + cos²x = 1, this becomes −sin x − 1 = −(1 + sin x).
5. So y′ = −(1 + sin x) / (1 + sin x)² = −1 / (1 + sin x).

Answer−1 / (1 + sin x)

Watch this explained “Cosine over sine, and the sign that flips”, 16:39 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (sin x + cos x) / (sin x − cos x). Use the quotient rule with u = sin x + cos x, v = sin x − cos x.
2. u′ = cos x − sin x, and v′ = cos x + sin x.
3. Numerator = u′v − uv′ = (cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x).
4. The first part is −(sin x − cos x)², and the second is −(sin x + cos x)².
5. Adding these: −[(sin x − cos x)² + (sin x + cos x)²] = −[2sin²x + 2cos²x] = −2.
6. So y′ = −2 / (sin x − cos x)².

Answer−2 / (sin x − cos x)²

Watch this explained “Cosine over sine, and the sign that flips”, 16:39 into Rules for differentiating a sum, a product and a quotient

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1. Write sec x = 1/cos x, so y = (1/cos x − 1) / (1/cos x + 1).
2. Multiply the top and bottom by cos x: y = (1 − cos x) / (1 + cos x).
3. Use the quotient rule with u = 1 − cos x (u′ = sin x) and v = 1 + cos x (v′ = −sin x).
4. Numerator = sin x(1 + cos x) − (1 − cos x)(−sin x) = sin x(1 + cos x) + sin x(1 − cos x) = 2 sin x.
5. So y′ = 2 sin x / (1 + cos x)².

Answer2 sin x / (1 + cos x)²

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Here n is an integer. First take n to be a positive whole number, so y = sinⁿx = sin x × sin x × … × sin x (n factors).
2. Try small cases with the product rule (uv)′ = u′v + uv′. For n = 2: (sin x · sin x)′ = cos x · sin x + sin x · cos x = 2 sin x cos x. For n = 3: (sin²x · sin x)′ = (2 sin x cos x) sin x + sin²x cos x = 3 sin²x cos x.
3. Each extra factor adds one more copy: if (sin^kx)′ = k sin^k−1x cos x, then (sin^k+1x)′ = (sin^kx · sin x)′ = k sin^k−1x cos x · sin x + sin^kx · cos x = (k + 1) sin^kx cos x.
4. So for every positive whole number n, y′ = n sinⁿ⁻¹x cos x.
5. For n = 0, y = 1 and y′ = 0, which the same formula also gives.
6. If n is negative, write n = −m with m a positive whole number, so y = 1 / sin^mx (wherever sin x ≠ 0). The quotient rule gives y′ = (0 × sin^mx − 1 × m sin^m−1x cos x) / sin^2mx = −m sin^−m−1x cos x = n sinⁿ⁻¹x cos x.
7. So for every integer n: y′ = n sinⁿ⁻¹x cos x.

Answern sinⁿ⁻¹x cos x

Watch this explained “Sine squared, by the rule”, 14:28 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (a + b sin x) / (c + d cos x). Use the quotient rule with u = a + b sin x, v = c + d cos x.
2. u′ = b cos x, and v′ = −d sin x.
3. Numerator = b cos x(c + d cos x) − (a + b sin x)(−d sin x) = bc cos x + bd cos²x + ad sin x + bd sin²x.
4. Since bd cos²x + bd sin²x = bd, the numerator becomes bd + ad sin x + bc cos x.
5. So y′ = (bd + ad sin x + bc cos x) / (c + d cos x)².

Answer(bd + ad sin x + bc cos x) / (c + d cos x)²

Watch this explained “Cosine over sine, and the sign that flips”, 16:39 into Rules for differentiating a sum, a product and a quotient

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1. Write y = sin(x + a) / cos x. Use the quotient rule with u = sin(x + a), v = cos x.
2. Since a is a fixed constant, u′ = cos(x + a) (treating x + a as the angle). Also v′ = −sin x.
3. Numerator = cos(x + a) cos x − sin(x + a)(−sin x) = cos(x + a) cos x + sin(x + a) sin x.
4. This matches the formula cos(A − B) = cos A cos B + sin A sin B with A = x + a, B = x, so it equals cos((x+a) − x) = cos a.
5. So y′ = cos a / cos²x.

Answercos a / cos²x

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Write y = x⁴(5 sin x − 3 cos x). Use the product rule with u = x⁴ (u′ = 4x³) and v = 5 sin x − 3 cos x (v′ = 5 cos x + 3 sin x).
2. y′ = 4x³(5 sin x − 3 cos x) + x⁴(5 cos x + 3 sin x) = 20x³ sin x − 12x³ cos x + 5x⁴ cos x + 3x⁴ sin x.
3. Take x³ common from every term: y′ = x³[(20 + 3x) sin x + (5x − 12) cos x].

Answerx³[(3x + 20) sin x + (5x − 12) cos x]

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (x² + 1) cos x. Use the product rule with u = x² + 1 (u′ = 2x) and v = cos x (v′ = −sin x).
2. y′ = 2x cos x + (x² + 1)(−sin x) = 2x cos x − (x² + 1) sin x.

Answer2x cos x − (x² + 1) sin x

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (ax² + sin x)(p + q cos x). Use the product rule with u = ax² + sin x (u′ = 2ax + cos x) and v = p + q cos x (v′ = −q sin x).
2. y′ = (2ax + cos x)(p + q cos x) + (ax² + sin x)(−q sin x).
3. y′ = (2ax + cos x)(p + q cos x) − q sin x (ax² + sin x).

Answer(2ax + cos x)(p + q cos x) − q sin x (ax² + sin x)

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (x + cos x)(x − tan x). Use the product rule with u = x + cos x (u′ = 1 − sin x) and v = x − tan x (v′ = 1 − sec²x = −tan²x, using sec²x − 1 = tan²x).
2. y′ = (1 − sin x)(x − tan x) + (x + cos x)(−tan²x).
3. Expand the first bracket: x − tan x − x sin x + sin x tan x. The second part is −x tan²x − cos x tan²x.
4. Since cos x tan²x = sin x tan x, the terms sin x tan x and −cos x tan²x cancel out.
5. What remains: y′ = x − tan x − x sin x − x tan²x.

Answerx − x sin x − x tan²x − tan x

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (4x + 5 sin x) / (3x + 7 cos x). Use the quotient rule (u/v)′ = (u′v − uv′) / v² with u = 4x + 5 sin x and v = 3x + 7 cos x.
2. u′ = 4 + 5 cos x, and v′ = 3 − 7 sin x.
3. So y′ = [(4 + 5 cos x)(3x + 7 cos x) − (4x + 5 sin x)(3 − 7 sin x)] / (3x + 7 cos x)².
4. First product: (4 + 5 cos x)(3x + 7 cos x) = 12x + 28 cos x + 15x cos x + 35 cos²x.
5. Second product: (4x + 5 sin x)(3 − 7 sin x) = 12x − 28x sin x + 15 sin x − 35 sin²x.
6. Subtract the second from the first: the 12x terms cancel, and 35 cos²x + 35 sin²x = 35(cos²x + sin²x) = 35.
7. Numerator = 28x sin x + 15x cos x − 15 sin x + 28 cos x + 35.
8. So y′ = (28x sin x + 15x cos x − 15 sin x + 28 cos x + 35) / (3x + 7 cos x)².

Answer(28x sin x + 15x cos x − 15 sin x + 28 cos x + 35) / (3x + 7 cos x)²

Watch this explained “Cosine over sine, and the sign that flips”, 16:39 into Rules for differentiating a sum, a product and a quotient

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1. cos(π/4) is the fixed number √2/2, so y = (√2/2) × x² / sin x.
2. Use the quotient rule on x² / sin x, with u = x² (u′ = 2x) and v = sin x (v′ = cos x).
3. (x² / sin x)′ = (2x sin x − x² cos x) / sin²x.
4. Multiplying by the constant √2/2: y′ = √2(2x sin x − x² cos x) / (2 sin²x).

Answer√2 (2x sin x − x² cos x) / (2 sin²x)

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Write y = x / (1 + tan x). Use the quotient rule with u = x (u′ = 1), v = 1 + tan x (v′ = sec²x).
2. y′ = [1 × (1 + tan x) − x × sec²x] / (1 + tan x)² = [(1 + tan x) − x sec²x] / (1 + tan x)².

Answer[(1 + tan x) − x sec²x] / (1 + tan x)²

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient

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1. Write y = (x + sec x)(x − tan x). Use the product rule with u = x + sec x (u′ = 1 + sec x tan x) and v = x − tan x (v′ = 1 − sec²x = −tan²x).
2. y′ = (1 + sec x tan x)(x − tan x) + (x + sec x)(−tan²x).
3. Expand the first bracket: x − tan x + x sec x tan x − sec x tan²x. The second part is −x tan²x − sec x tan²x.
4. Adding these together: y′ = x − tan x + x sec x tan x − x tan²x − 2 sec x tan²x.

Answerx − tan x + x sec x tan x − x tan²x − 2 sec x tan²x

Watch this explained “Written short”, 7:21 into Rules for differentiating a sum, a product and a quotient

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1. Write y = x / sinⁿx. Use the quotient rule with u = x (u′ = 1) and v = sinⁿx (v′ = n sinⁿ⁻¹x cos x, from the earlier rule for sinⁿx).
2. y′ = [sinⁿx − x × n sinⁿ⁻¹x cos x] / sin²ⁿx.
3. Take sinⁿ⁻¹x common from the numerator: sinⁿ⁻¹x(sin x − nx cos x).
4. Cancel sinⁿ⁻¹x from top and bottom: y′ = (sin x − nx cos x) / sinⁿ⁺¹x.

Answer(sin x − nx cos x) / sinⁿ⁺¹x

Watch this explained “The quotient rule”, 8:10 into Rules for differentiating a sum, a product and a quotient