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प्रश्न
Find the derivative of the following function:
3cot x + 5cosec x
उत्तर
(x) = cot x
f(x + h) = cot (x + h)
f(x + h) – f(x) = cot (x + h) – cot x
= `(cos (x +h))/(sin (x + h)) - cos x/sin x`
= `(cos (x +h) sin x - cos x sin (x +h))/(sin (x +h) sin x)`
= `(-[sin (x +h) cosx - cos(x +h) sin x])/(sin (x +h) sin x)`
= `(sin (x + h - x))/(sin (x + h)sin x)`
= `- sin h/(sin (x +h) sin x)`
`f'(x) = lim_(h → 0)(f(x +h) - f(x))/h`
= ` lim_(h → 0) 1/h. [sin h/(sin (x +h)sin x)]`
= `-1. 1/(sin x sin x)`
= `- cosec^2 x`
Now, f'(x)= `3 d/dx cotx + 5 d/dx cosec x`
= 3(- cosec2 x) + 5 (- cosec x.cot x)
= – 3 cosec2x – 5 cosec x cot x
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