Exercise 12.2 answers: Limits and Derivatives

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1. By first principle, the derivative of f(x) at a point x is the limit as h → 0 of [f(x+h) − f(x)]/h.
2. Here f(x) = x² − 2, so f(x+h) − f(x) = (x+h)² − 2 − (x² − 2) = 2xh + h².
3. Divide by h: 2x + h, and let h → 0 to get 2x.
4. So the derivative of x² − 2 is 2x, and at x = 10 this is 2 × 10 = 20.

Answer20

Watch this explained “What the definition delivers”, 7:58 into The rate of change at a point, defined as a limit of average rates

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1. By first principle, [f(x+h) − f(x)]/h with f(x) = x gives [(x+h) − x]/h = h/h = 1.
2. This is already 1 for every h ≠ 0, so the limit as h → 0 is 1.
3. So the derivative of x is 1 everywhere, in particular at x = 1.

Answer1

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1. By first principle, [f(x+h) − f(x)]/h with f(x) = 99x gives [99(x+h) − 99x]/h = 99h/h = 99.
2. This is 99 for every h ≠ 0, so the limit as h → 0 is 99.
3. So the derivative of 99x is 99 everywhere, in particular at x = 100.

Answer99

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(i) x³−27

1. Let f(x) = x³ − 27. By first principle, f′(x) = lim (h→0) [f(x + h) − f(x)]/h.
2. f(x + h) − f(x) = (x + h)³ − 27 − (x³ − 27) = 3x²h + 3xh² + h³.
3. Divide by h: 3x² + 3xh + h². Now let h → 0 to get 3x².

Answer3x²

(ii) (x−1)(x−2)

1. First multiply out: f(x) = (x − 1)(x − 2) = x² − 3x + 2.
2. f(x + h) − f(x) = [(x + h)² − 3(x + h) + 2] − [x² − 3x + 2] = 2xh + h² − 3h.
3. Divide by h: 2x + h − 3. Now let h → 0 to get 2x − 3.

Answer2x − 3

(iii) 1/x²

1. Let f(x) = 1/x². Then f(x + h) − f(x) = 1/(x + h)² − 1/x² = [x² − (x + h)²] / [x²(x + h)²].
2. The top is x² − (x² + 2xh + h²) = −2xh − h² = −h(2x + h).
3. Divide by h: −(2x + h) / [x²(x + h)²]. Now let h → 0 to get −2x / (x² · x²) = −2/x³.

Answer−2/x³

(iv) (x+1)/(x−1)

1. Let f(x) = (x + 1)/(x − 1). Then f(x + h) − f(x) = (x + h + 1)/(x + h − 1) − (x + 1)/(x − 1).
2. Over the common denominator (x + h − 1)(x − 1), the top is (x + h + 1)(x − 1) − (x + 1)(x + h − 1).
3. (x + h + 1)(x − 1) = x² + hx − h − 1 and (x + 1)(x + h − 1) = x² + hx + h − 1, so the top is −2h.
4. Divide by h: −2 / [(x + h − 1)(x − 1)]. Now let h → 0 to get −2/(x − 1)².

Answer−2/(x − 1)²

Watch this explained “Taking the factor out”, 3:40 into The rate of change at a point, defined as a limit of average rates

Open NCERT p. 248One way to think about it

1. Given: f(x) = x¹⁰⁰/100 + x⁹⁹/99 + … + x²/2 + x + 1. Prove: f′(1) = 100 f′(0).
2. Differentiate term by term: the derivative of xᵏ/k is xᵏ⁻¹, for each k from 100 down to 2.
3. The derivative of the x term is 1, and the derivative of the constant 1 is 0.
4. So f′(x) = x⁹⁹ + x⁹⁸ + ... + x + 1 (100 terms, from x⁹⁹ down to the constant 1).
5. At x = 0, every term with a power of x becomes 0, leaving only the last term: f′(0) = 1.
6. At x = 1, every power of 1 is 1, so f′(1) is just 100 ones added up: f′(1) = 100.
7. So f′(1) = 100 = 100 × 1 = 100 f′(0), which is what we had to prove.

In shortf′(1) = 100 and f′(0) = 1, so f′(1) = 100 f′(0).

Watch this explained “The exponent cancels its own division”, 12:37 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. Differentiate: xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + … + aⁿ⁻¹x + aⁿ.
2. The function is a sum of terms aᵏxⁿ⁻ᵏ, for k = 0, 1, ..., n (a is just a fixed number, like a coefficient).
3. Differentiate each term with the power rule: the derivative of aᵏxⁿ⁻ᵏ is aᵏ(n−k)xⁿ⁻ᵏ⁻¹.
4. The very last term, aⁿ (when k = n), has no x in it, so its derivative is 0 and it drops out.
5. Adding up the rest gives the derivative: nxⁿ⁻¹ + (n−1)axⁿ⁻² + (n−2)a²xⁿ⁻³ + ... + aⁿ⁻¹.

Answernxⁿ⁻¹ + (n − 1)axⁿ⁻² + (n − 2)a²xⁿ⁻³ + ... + aⁿ⁻¹

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(i) (x − a)(x − b)

1. Multiply out (x − a)(x − b) = x² − (a + b)x + ab.
2. Differentiate term by term: the derivative of x² is 2x, of −(a + b)x is −(a + b), and the constant ab gives 0.

Answer2x − (a + b)

(ii) (ax² + b)²

1. Square the bracket: (ax² + b)² = a²x⁴ + 2abx² + b².
2. Differentiate term by term: 4a²x³ + 4abx.

Answer4a²x³ + 4abx

(iii) (x − a)/(x − b)

1. This is one bracket divided by another, so use the quotient rule: (u/v)′ = (u′v − uv′)/v².
2. Here u = x − a gives u′ = 1, and v = x − b gives v′ = 1.
3. u′v − uv′ = (x − b) − (x − a) = a − b, so the derivative is (a − b)/(x − b)².

Answer(a − b)/(x − 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: (xⁿ − aⁿ)/(x − a).
2. This is one function over another, so use the quotient rule: (u/v)′ = (u′v − uv′)/v².
3. Here u = xⁿ − aⁿ, so u′ = nxⁿ⁻¹ (a is a fixed constant, so aⁿ differentiates to 0).
4. And v = x − a, so v′ = 1.
5. Put these into the quotient rule and leave the answer over v² = (x − a)².

Answer[nxⁿ⁻¹(x − a) − (xⁿ − aⁿ)]/(x − a)²

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(i) 2x − 3/4

1. The derivative of 2x is 2, and the derivative of the constant −3/4 is 0.

Answer2

(ii) (5x³ + 3x − 1)(x − 1)

1. Multiply out (5x³ + 3x − 1)(x − 1) first.
2. 5x³·x − 5x³·1 + 3x·x − 3x·1 − 1·x + 1·1 gives 5x⁴ − 5x³ + 3x² − 3x − x + 1.
3. Collect like terms: 5x⁴ − 5x³ + 3x² − 4x + 1.
4. Differentiate term by term: 20x³ − 15x² + 6x − 4.

Answer20x³ − 15x² + 6x − 4

(iii) x⁻³(5 + 3x)

1. Multiply out x⁻³(5 + 3x) = 5x⁻³ + 3x⁻².
2. Differentiate each power: 5x⁻³ gives −15x⁻⁴, and 3x⁻² gives −6x⁻³.

Answer−15/x⁴ − 6/x³

(iv) x⁵(3 − 6x⁻⁹)

1. Multiply out x⁵(3 − 6x⁻⁹) = 3x⁵ − 6x⁻⁴.
2. Differentiate each power: 3x⁵ gives 15x⁴, and −6x⁻⁴ gives 24x⁻⁵.

Answer15x⁴ + 24/x⁵

(v) x⁻⁴(3 − 4x⁻⁵)

1. Multiply out x⁻⁴(3 − 4x⁻⁵) = 3x⁻⁴ − 4x⁻⁹.
2. Differentiate each power: 3x⁻⁴ gives −12x⁻⁵, and −4x⁻⁹ gives 36x⁻¹⁰.

Answer−12/x⁵ + 36/x¹⁰

(vi) 2/(x + 1) − x²/(3x − 1)

1. Differentiate the two parts separately and subtract.
2. For 2/(x + 1), use the quotient rule with u = 2, v = x + 1: this gives −2/(x + 1)².
3. For x²/(3x − 1), use the quotient rule with u = x², v = 3x − 1: u′v − uv′ = 2x(3x − 1) − x²(3) = 3x² − 2x, over (3x − 1)².
4. Subtract the second result from the first.

Answer−2/(x + 1)² − (3x² − 2x)/(3x − 1)²

Watch this explained “Multiply out first”, 15:06 into The power rule, and a polynomial's derivative assembled out of it and the sum rule

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1. By first principle, f′(x) = lim(h→0) [cos(x + h) − cos x]/h.
2. Use the identity cos(x + h) − cos x = −2 sin(x + h/2) sin(h/2), which turns the difference into a product.
3. Write the quotient as −sin(x + h/2) · [sin(h/2)/(h/2)].
4. As h→0, sin(x + h/2) → sin x, and sin(h/2)/(h/2) → 1.
5. So the derivative is −sin x.

Answer−sin x

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(i) sin x cos x

1. Use the product rule on sin x · cos x: (uv)′ = u′v + uv′.
2. u = sin x has u′ = cos x, and v = cos x has v′ = −sin x.
3. So the derivative is cos x · cos x + sin x · (−sin x) = cos²x − sin²x.

Answercos²x − sin²x

(ii) sec x

1. Write sec x = 1/cos x and use the quotient rule.
2. u = 1 gives u′ = 0; v = cos x gives v′ = −sin x.
3. (u′v − uv′)/v² = (0 − (−sin x))/cos²x = sin x/cos²x = sec x tan x.

Answersec x tan x

(iii) 5 sec x + 4 cos x

1. Differentiate each term separately and add.
2. The derivative of 5 sec x is 5 sec x tan x, and the derivative of 4 cos x is −4 sin x.

Answer5 sec x tan x − 4 sin x

(iv) cosec x

1. Write cosec x = 1/sin x and use the quotient rule.
2. u = 1 gives u′ = 0; v = sin x gives v′ = cos x.
3. (u′v − uv′)/v² = (0 − cos x)/sin²x = −cos x/sin²x = −cosec x cot x.

Answer−cosec x cot x

(v) 3 cot x + 5 cosec x

1. Differentiate each term separately and add.
2. The derivative of 3 cot x is −3 cosec²x, and the derivative of 5 cosec x is −5 cosec x cot x.

Answer−3 cosec²x − 5 cosec x cot x

(vi) 5 sin x − 6 cos x + 7

1. Differentiate each term: 5 sin x gives 5 cos x, −6 cos x gives 6 sin x, and the constant 7 gives 0.

Answer5 cos x + 6 sin x

(vii) 2 tan x − 7 sec x

1. Differentiate each term separately: 2 tan x gives 2 sec²x, and −7 sec x gives −7 sec x tan x.

Answer2 sec²x − 7 sec x tan x

Watch this explained “The rest of the family”, 11:07 into Sine and tangent go back to the definition, because no rule so far reaches them