Question

cot x differentation of ist principle solve

Answer 1

Hi,

Here is your answer,

Cot x differentiation by using 1st Principle

y = cotx ; d/dx(cotx) = -cosec²x

Let y = f(x) = cot x

f(x+h) = cot(x+h)

f'(x) = $\lim_{h ->0}$ f(x+h) - f(x) / h

= $\lim_{h->0}$ cot(x+h) - cot(x) / h

= $\lim_{h->0}$ cosx (x+h)/sin(x+h) - cosx/sinx / h

= $\lim_{h->0}$ 1/h [cos(x+h)/sin(x+h) - cosx/sinx]

= $\lim_{h->0}$ cos(x+h). sinx. cosx. sin(x+h) ( B (cos(x+h) sinx(A) cosx(A) sin(x+h)(B)

= $\lim_{h->0}$ sin(x-(x+h)/sin(x+h). sinx (∴ +x and -x will get cancel)

= $\lim_{h->0}$ sin(-h)/sin(x+h).sinx

= $\lim_{h->0}$ sin h/ h sin(x+h). sinx { sin(θ) = sinθ }

= $\lim_{h->0}$ sin h/h.sin(x+h). sin(x)

= -1/sinx. $\lim_{h->0}$ sin h/h.sin(x+h)

= -1/sinx [$\lim_{h->0}$ sinh/h] $\lim_{h->0}$(1/sin(x+h)

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

= -1/sinx . 1/sinx

= -1/sinx

Therefore, d/dx(cotx) = -cosec³x

NOTE:- (.) THIS DOT INDICATES ''×''(MULTIPLY) SYMBOL.

Hope it helps you !