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.Misc. Q.1

Find the derivative of the following functions from first principles:

1. −x

2. (-x)^-1

3. sin (x + 1)

4. cos (x – )

1.Let f(x) = -x

Now, by first principle, f’(x) = lim_h->0 [{f (x + h) – f(x)} ÷ h]

= lim_h->0 [{- (x + h) – (-x)} ÷ h]

= lim_h->0 [{-x - h + h} ÷ h]

= lim_h->0 [-h ÷ h]

= lim_h->0 (-1)

= -1

2. Let f(x) = (-x)^-1 = -1 ÷ (-x) = -

Now, by first principle, f’(x) = lim_h->0 [{f (x + h) – f(x)} ÷ h]

= lim_h->0 [{-1 ÷ (x + h) – (- )} ÷ h]

= lim_h->0 [{-1 ÷ (x + h) + } ÷ {h × x (x + h)}]

= lim_h->0 [(-x + x + h) ÷ {h × x (x + h)}]

= lim_h->0 [h ÷ {h × x (x + h)}]

= lim_h->0 [1 ÷ {x (x + h)}]

= 1 ÷ (x × x)

= (1 ÷ x²)

3.Let f(x) = sin (x + 1)

Now, by first principle, f’(x) = lim_h->0 [{f (x + h) – f(x)} ÷ h]

= lim_h->0 [{sin (x + h + 1) – sin (x + 1)} ÷ h]

= lim_h->0 [{2 × cos (x + h + 1 + x + 1) ÷ 2 × sin (x + h + 1 – x - 1) ÷ 2} ÷ h]

= lim_h->0 [{(2 × cos (2x + h + 2) ÷ 2) × sin ( )} ÷ h]

= lim_h->0 [{cos (2x + h + 2) ÷ 2} × {sin ( ) ÷ ( )}]

= lim_h->0 [cos (2x + h + 2) ÷ 2] × lim_h->0 [sin ( )÷( )]

= {cos (2x + 0 + 2) ÷ 2} × 1

[Since lim_x->0 (sin x) ÷ x = 1]

= cos (x + 1)

4. Let f(x) = cos (x – ) Now, by first principle,

f’(x) = lim_h->0 [{f (x + h) – f(x)} ÷ h]

= lim_h->0 [{cos (x + h - ) – cos (x - )} ÷ h]

= lim_h->0 [{-2 × sin (x + h - + x - ) ÷ 2 × sin (x + h - - x + ) ÷ 2} ÷ h]

= lim_h->0 [{-2 × sin (2x + h - ) ÷ {2 × sin ( )} ÷ h}]

= -lim_h->0 [sin (2x + h - ) ÷ 2 × sin ( ) ÷ ( )]

= -lim_h->0 [sin (2x + h - ) ÷ 2] × lim_h->0 [sin ( ) ÷ ( )]

= {-sin (2x + 0 - ) ÷ 2} × 1

= -sin (x - )

Limits and Derivatives Class 11 Maths NCERT Chapter 12 NCERT Solutions

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