Question

Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

cosec x cot x

Answer 1

Let f(x) = cosec x cot x

By Leibnitz product rule,

f'(x) = cosec x (cot x)' + cot x (cosec x)'      ...(1)

Let f (x) = cot x. Accordingly, f(x + h) = cot (x + h)

By first principle,

Let f₁(x) = `lim_(h->0) (f_1(x + h)− f_1(x))/h`

= `lim_(h->0) ((cot (x + h) -cot x)/h)`

= `lim_(h->0) (cos (x + h)/(sin (x + h))-(cos x)/(sin x))`

= `lim_(h->0)1/h[(sin x cos (x + h) - cos x sin (x + h))/(sin x sin (x + h))]`

= `lim_(h->0)1/h[sin (x - h - h)/(sin x sin (x + h))]`

= `1/(sin x) lim_(h->0)1/h[sin (- h)/(sin (x + h))]`

= `1/(sin x) (lim_(h->0) (sin h)/h) (lim_(h->0) 1/(sin (x + h)))`

= `-1/(sin x).1 (1/(sin (x + 0)))`

= `(-1)/(sin^2 x)`

= - cosec² x

∴ (cot x)' = - cosec² x       ...(2)

Now, let f₂(x) = cosec x. Accordingly, f₂(x + h) = cosec(x + h)

By first principle,

f₂(x)' = `lim_(h->0) (f_2 (x + h) - f_2 (x))/h`

= `lim_(h->0) 1/h [cosec (x + h) - cosec x]`

= `lim_(h->0)1/h [1/(sin (x + h)) - 1/(sin x)]`

= `lim_(h->0)1/h [(sin x - sin (x + h))/(sin x sin (x + h))]`

= `1/(sin x). lim_(h->0)1/h[(2 cos  ((x + x + h)/2) sin  ((x - x - h)/2))/(sin (x + h))]`

= `1/(sin x). lim_(h->0)1/h [(2 cos  ((2x + h)/2) sin  ((-h)/2))/(sin (x + h))]`

 

= `1/(sin x). lim_(h->0)1/h [(2 cos  ((2x + h)/2) sin  ((-h)/2))/(sin (x + h))]`

= `1/sin x. lim_(h->0) [-sin(h/2)/((h/2)) (cos ((2x +h)/2))/(sin (x + h))]`

= `(-1)/(sin x). lim_(h->0) sin(h/2)/((h/2)) lim_(h->0) (cos  ((2x + h)/2))/(sin (x + h))`

= `(-1)/ (sin x).1 (cos((2x + 0)/2))/(sin (x + 0)`

= `(-1)/(sin x).(cos x)/(sin x)`

= -cosecx . cot x

∴ (cosec x) = -cosec x. cot x      ...(3)

From (1), (2), and (3), we obtain

f'(x) = cosec x(-cosec²x) + cot x (-cosec x cot x)

= -cosec³ x-cot² x cosec x