Find the derivative of the following functions from first principle.
(i) x³ - 27   (ii) (x - 1)(x - 2)   (iii) 1/x²   (iv) (x + 1)/(x - 1)

Solution:

(i) Let f (x) = x3 - 27

Accordingly, from the first principle,

f' (1) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [(x + h)3 - 27] - (x3 - 27)] / h

= limₕ→₀ (x3 + h3 + 3x2h + 3xh2 - 27 - x3 + 27) / h

= limₕ→₀ (h3 + 3x2h + 3xh2)/h

= limₕ→₀ h(h2 + 3x2 + 3xh) / h

= limₕ→₀ (h2 + 3x2 + 3xh)

= 0 + 3x2 + 0

= 3x2

(ii) Let f (x) = (x - 1)(x - 2)

Accordingly, from the first principle

f' (1) = 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 - 3

(iii) Let f (x) = 1/x2

Accordingly, from the first principle

f' (1) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [1/(x + h)2 - 1/x2]/h

= limₕ→₀ 1(/h) [x2 - (x + h)2] / (x2 (x + h)2)]

= limₕ→₀ (1/h) (x2 - x2 - 2hx - h2) / (x2 (x + h)2)

= limₕ→₀ (1/h) (- h2 - 2hx) / (x2 (x + h)2)

= limₕ→₀ (- h - 2x) / (x2 (x + h)2)

= (0 - 2x)/(x2 (x + 0)2)

= - 2/x3

(iv) Let f (x) = (x + 1)/(x - 1)

Accordingly, from the first principle

f' (1) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [(x + h + 1)/(x + h - 1) - (x + 1)/(x - 1)]/h

= limₕ→₀ (1/h) [(x - 1)(x + h + 1) - (x + 1)(x + h - 1)] / [( x - 1)(x + h - 1)]

= limₕ→₀ (1/h) [(x2 + hx + x - x - h - 1) - (x2 + hx - x + x + h - 1)] / [(x - 1)( x + h - 1)]

= limₕ→₀ (1/h) (- 2h) / [(x - 1)(x + h - 1)]

= limₕ→₀ (- 2) / [(x - 1)(x + h - 1)]

= (- 2) / [( x - 1)( x - 1)]

= - 2/(x - 1)2

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.2 Question 4

Find the derivative of the following functions from first principle. (i) x³ - 27 (ii) (x - 1)(x - 2) (iii) 1/x² (iv) (x + 1)/(x - 1)

Summary:

The derivatives of the given functions are (i) 3x2 (ii) 2x - 3 (iii) - 2/x3 (iv) - 2/(x - 1)2

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