. derivative of cotx by first principle

Best Answer

Derivation of Cot x by First Principle:

Let us take f(x) = cot x.

By the first principle, the derivative of f(x) is given by the following limit.

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

Since f(x) = cot x,

So, f (x + h) = cot (x + h).

Substituting these in (1),

f'(x) = limₕ→₀ [cot (x + h) - cot x] / h

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

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

We have, sin A cos B - cos A sin B = sin (A - B).

f'(x) = limₕ→₀ [ sin (x - (x + h))] / [ h sin x · sin (x + h)]

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

We have sin (-h) = - sin h.

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

By limit formulas, limₕ→₀ (sin h)/ h = 1.

f'(x) = -1 [ 1 / (sin x · sin (x + 0))] = -1/sin2x

We know that the reciprocal of sin is csc. So,

f'(x) = -csc²x.

Hence proved.