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प्रश्न
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
उत्तर
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
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