NCERT Solutions
Class 11 Maths
Limits and Derivatives
Ex.13.2 Q.4
Find the derivative of the following functions from first principle.
(1) x3 – 27
(2) (x – 1) (x – 2)
(3) 1 ÷ x2
(4) (x + 1) ÷ (x - 1)
(1) Let f(x) = x3 – 27
Now, from the first principle, f’(x) = limh->0 [{f (x + h) – f(x)} ÷ h]
= limh->0 [{(x + h)3 – 27} – (x3 – 27)} ÷ h]
= limh->0 [{x3 + h3 + 3x2 h + 3xh2 – 27 – x3 + 27} ÷ h]
= limh->0 [{h3 + 3x2 h + 3xh2} ÷ h]
= limh->0 [h2 + 3x2 + 3xh]
= 0 + 3x2 + 3x × 0
= 3x2
(2) Let f(x) = (x – 1) (x – 2)
Now, from the first principle, f’(x) = limh->0 [{f (x + h) – f(x)} ÷ h]
= limh->0 [{(x + h – 1) (x + h – 2) – (x – 1) (x – 2)} ÷ h]
= limh->0 [{(x2 + hx – 2x + hx + h2 – 2h – x – h + 2) (x2 - 2x – x + 2)} ÷ h]
= limh->0 [{2hx + h2 – 3h} ÷ h]
= limh->0 [2x + h - 3]
= 2x + 0 - 3
= 2x – 3
(3) Let f(x) = (1 ÷ x2)
Now, from the first principle, f’(x) = limh->0 [{f (x + h) – f(x)} ÷ h]
= limh->0 [{(1 ÷ (x + h)2) – (1 ÷ x2)} ÷ h]
= limh->0 [{(x2) - (x + h)2} ÷ {h (x + h)2 x2}]
= limh->0 [{x2 - x2 - h2 – 2hx)} ÷ {h (x + h)2 x2}]
= limh->0 [{h2 – 2hx)} ÷ {h (x + h)2 x2}]
= limh->0 [{h – 2x)} ÷ {(x + h)2 x2}]
= limh->0 [{0 – 2x)} ÷ {(x + 0)2 x2}]
= -(2x ÷ x4)
= -(2 ÷ x3)
(4) Let f(x) = (x + 1) ÷ (x - 1)
Now, from the first principle,
f’(x) = limh->0 [{f (x+ h) – f(x)} ÷ h]
= limh->0 [{(x + h + 1) ÷ (x + h - 1) – (x + 1) ÷ (x – 1)} ÷ h]
= limh->0 [{(x - 1) (x + h + 1) – (x + 1) (x + h - 1)} ÷ {h (x - 1) ÷ (x + h – 1)}]
= limh->0 [{(x2 + hx + x – x - h - 1) – (x2 + hx - x + x + h - 1)} ÷ {h (x - 1) ÷ (x + h – 1)}]
= limh->0 [(-2h) ÷ {h (x - 1) ÷ (x + h – 1)}]
= limh->0 [(-2) ÷ {(x - 1) ÷ (x + h – 1)}]
= (-2) ÷ {(x - 1) ÷ (x + 0 – 1)}
= -2 ÷ {(x - 1) (x - 1)}
= -2 ÷ (x - 1)2