What is the derivative of y=cot(x) using the first principal rule?

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Jim H Share
Nov 20, 2016

Answer:

See the explanation.

Explanation:

Let #f(x) = cotx#, then , by definition, #f'(x) = lim_(hrarr0)(cot(x+h)-cotx)/h#.

Well try to simplify using #cottheta = cos theta/sintheta# and also using the sum laws for sine and for cosine.

#f'(x) = lim_(hrarr0)(cot(x+h)-cotx)/h#

# = lim_(hrarr0)(cos(x+h)/sin(x+h)-cosx/sinx)/h#

# = lim_(hrarr0)(sinxcos(x+h) - cosxsin(x+h))/(hsin(x+h)sinx#

# = lim_(hrarr0)(sinx[cosxcos h-sinxsin h] - cosx[sinxcos h+cosxsin h])/(hsin(x+h)sinx#

# = lim_(hrarr0)(-sin^2xsin h - cos^2xsin h)/(hsin(x+h)sinx#

# = lim_(hrarr0)(-sin h)/(hsin(x+h)sinx#

# = lim_(hrarr0)-sin h/h 1/(sin(x+h)sinx)#

# = -(1)1/(sinxsinx) = -1/sin^2x = -csc^2x#

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