find the derivative of the function cot x from the first principle
Answers
Refer to the attachment....
Answer:
The derivative of the function cot x = - cosec^2(x)
Solution:
f'(x) = limₕ→₀ [f(x + h) - f(x)] / h
Since f(x) = cot x, we have 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
By sum and difference formulas, 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 cosec. So,
f'(x) = -cosec2x.
Hence proved.