NCERT solution Class 11 limits and derivatives

NCERT Solutions

Class 11 Maths

Limits and Derivatives

NCERT Questions

Ex.13.1 Q.1
Ex.13.1 Q.2
Ex.1.31 Q.3
Ex.13.1 Q.4
Ex.13.1 Q.5
Ex.13.1 Q.6
Ex.13.1 Q.7
Ex.13.1 Q.8
Ex.13.1 Q.9
Ex.13.1 Q.10
Ex.13.1 Q.11
Ex.13.1 Q.12
Ex.13.1 Q.13
Ex.13.1 Q.14
Ex.13.1 Q.15
Ex.13.1 Q.16
Ex.13.1 Q.17
Ex.13.1 Q.18
Ex.13.1 Q.19
Ex.13.1 Q.20
Ex.13.1 Q.21
Ex.13.1 Q.22
Ex.13.1 Q.23
Ex.13.1 Q.24
Ex.13.1 Q.25
Ex.13.1 Q.26
Ex.13.1 Q.27
Ex.13.1 Q.28
Ex.13.1 Q.29
Ex.13.1 Q.30
Ex.13.1 Q.31
Ex.13.1 Q.32
Ex.13.2 Q.1
Ex13.2 Q.3
Ex.13.2 Q.4
Ex.13.2 Q.5
Ex.13.2 Q.6
Ex.13.2 Q.7
Ex.13.2 Q.8
Ex.13.2 Q.9
Ex.13.2 Q.10
Ex.13.2 Q.11
Ex.Misc. Q.1
Ex.Misc. Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10
Ex.Misc.Q.11
Ex.Misc.Q.12
Ex.Misc.Q.13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.13
Ex.Misc.Q.17
Ex.Misc.Q.18
Ex.Misc. Q.19
Ex.Misc.Q.20
Ex.Misc.Q.21
Ex.Misc.Q.22
Ex.Misc. Q.23
Ex.Misc.Q.24
Ex.Misc.Q.25
Ex.Misc.Q.26
Ex.Misc.Q.27
Ex.Misc.Q.28
Ex.Misc.Q.29
Ex.Misc.Q.30

Ex.13.2 Q.9

Find the derivative of

(1) 2x –

(2) (5x³ + 3x – 1) (x – 1)

(3) x^–3(5 + 3x)

(4) x⁵(3 – 6x^–9)

(5) x^–4 (3 – 4x^–5)

(6) {2 ÷ (x + 1)} – {x² ÷ (3x - 1)}

(1) Let y = 2x –

Now, = d (2x – ) ÷ dx

=> = {d(2x) ÷ dx} - {d ( ) ÷ dx}

=> = 2 – 0

=> = 2

(2) Let y = (5x³ + 3x – 1) (x – 1)

Now, = d {(5x³ + 3x – 1) (x – 1)} ÷ dx

=> = (5x³ + 3x – 1) × (d (x – 1) ÷ dx) + {(x - 1) × d (5x³ + 3x – 1) ÷ dx}

=> = (5x³ + 3x – 1) × 1 + (x - 1) × (3 × 5x² + 3)

=> = (5x³ + 3x – 1) + (x - 1) (15x² + 3)

=> = 5x³ + 3x – 1 + 15x³ + 3x - 15x² – 3

=> = 20x³ - 15x² + 6x – 4

(2) Let y = x^–3(5 + 3x)

=> y = 5x^–3 + 3x^–3+1

=> y = 5x^–3 + 3x^–2

Now = {d (5x^–3 + 3x^–2) ÷ dx}

=> = {d(5x^–3) ÷ dx} + {d(3x^–2) ÷ dx}

=> = (-3) × 5x^–3-1 + (-2) × 3x^–2-1

=> = -15x^–4 - 6x^–3

=> = (-15 ÷ x⁴) – (6 ÷ x³)

=> = {-3(5 + 2x) ÷ x⁴}

(3) Let y = x⁵(3 – 6x^–9)

=> y = 3x⁵ – 6x^{–9 + 5}

=> y = 3x⁵ – 6x^–4

Now, = d (3x⁵ – 6x^–4) ÷ dx

=> = {d(3x⁵) ÷ dx} – {d(6x^–4) ÷ dx}

=> = 5 × 3x^5-1 – (-4) × 6x^–4-1

=> = 15x⁴ + 24x^–5

=> = 15x⁴ + (24 ÷ x⁵)

(4) Let y = x^–4 (3 – 4x^–5)

=> y = 3x^–4 – 4x^–5-4

=> y = 3x^–4 – 4x^–9

Now, = d (3x^–4 – 4x^–9) ÷ dx

=> = {d(3x^–4) ÷ dx} – {d(4x^–9) ÷ dx}

=> = (-4) × 3x^–4-1 – (-9) × 4x^–9-1

=> = -12x^–5 + 36x^–10

=> = (-12 ÷ x⁵) + (36 ÷ x¹⁰)

(5) Let y = {2 ÷ (x + 1)} – {x² ÷ (3x - 1)}

Now, = d {(2 ÷ (x + 1)) – (x² ÷ (3x - 1))} ÷ dx

=> = d {({2 ÷ (x + 1)}} ÷ dx) – d {x²÷ (3x - 1)} ÷ dx

=> = [{(x + 1) × (d (2) ÷ dx) – 2 × (d (x + 1) ÷ dx)} ÷ (x + 1)²]

- [{((3x - 1) × d(x²) ÷ dx) – {x² × d (3x - 1) ÷ dx}} ÷ (3x - 1)²]

=> = [{(x + 1) × 0 – 2 × 1} ÷ (x + 1)²] - [{(3x - 1) × 2x – 3x²} ÷ (3x - 1)²]

=> = {-2 ÷ (x + 1)²} - [{6x² – 2x - 3x²} ÷ (3x - 1)²]

=> = {-2 ÷ (x + 1)²} – {(3x² – 2x) ÷ (3x - 1)²}

=> = {-2 ÷ (x + 1)²} – {x (3x – 2) ÷ (3x - 1)²}

Limits and Derivatives Class 11 Maths NCERT Chapter 12 NCERT Solutions

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