Find the derivative of the following functions from first principle:
(i) -x (ii) (-x)⁻¹ (iii) sin (x + 1) (iv) cos (x - π/8)

Solution:

We know that the derivative of a function y = f(x) from the first principle is, f' (x) = limₕ→₀ [f (x + h) - f (x)]/h.

(i) The given function is f(x) = -x.

Its derivative is,

f' (x) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [ -(x+h) - (-x)] / h

= limₕ→₀ [-x - h + x] / h

= limₕ→₀ [-h] / h

= limₕ→₀ -1

= -1

(ii) The given function is f(x) = (-x)⁻¹ = 1 / (-x) = - 1/x. Its derivative is,

f' (x) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [ -1/(x+h) - (-1/x) ] / h

= limₕ→₀ [ -1 / (x+h) + 1/x ] / h

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

= limₕ→₀ [ h / (x(x+h)) ] · (1/h)

= limₕ→₀ [ 1/ (x(x+h)) ]

= 1/(x(x+0))

= 1/x²

(iii) The given function is f(x) = sin (x + 1). Its derivative is,

f' (x) = limₕ→₀ [f (x + h) - f (x)]/h

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

We know that sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2

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

= limₕ→₀ [ 2 cos ((2x+h+2)/2) sin (h/2) ] / h

Multiply and divide by 1/2,

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

= [2 cos ((2x+2)/2)] [(1/2) (1)] (As limₓ→₀ sin x / x = 1)

= cos (x + 1)

(iv) The given function is f(x) = cos (x - π/8). Its derivative is,

f' (x) = limₕ→₀ [f (x + h) - f (x)]/h

= limₕ→₀ [ cos (x + h - π/8) - cos (x - π/8) ] / h

We know that cos A - cos B = - 2 sin (A+B)/2 sin (A-B)/2

= limₕ→₀ [ - 2 sin ((x+h-π/8 + x -π/8)/2) sin ((x+h-π/8 - x + π/8)/2) ] / h

= limₕ→₀ [ -2 sin ((2x + h -π/4)/2) sin (h/2) ] / h

Multiply and divide by 2,

= limₕ→₀ [ -2 sin ((2x + h -π/4)/2) [ (1/2) limₕ/₂→₀ sin (h/2) / (h/2)]

= -2 sin ((2x - π/4) /2) [(1/2)(1)] (As limₓ→₀ sin x / x = 1)

= - sin (x - π/8)

NCERT Solutions Class 11 Maths Chapter 13 Exercise ME Question 1

Solution:

The derivatives of the given functions from first principle are

(i) -1

(ii) 1/x²

(iii) cos (x + 1)

(iv) - sin (x - π/8)

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