Question

Find the derivate by first principal at any point x in f(x) =cos ec²x

Answer 1

Answer:

-[2cotx.cosec^2x]

Step-by-step explanation:

f(x)=cosec^2x

f'(x)=[f(x+h)-f(x)]/h

f'(x)=[cosec^2(x+h)-cosec^2x]/h

=[ (1/sin^2(x+h))-(1/sin^2x)]/h

taking LCM

= [sin^2x-sin^2(x+h)]/h(sin^2(x+h))(sin^2x)

since sin^2a-sin^2b=sin(a+b).sin(a-b)

[sin(2x+h).sin(-h)]/h(sin^2(x+h))(sin^2x)

now Sinh/h=1

-[sin(2x+h)]/sin^2(x+h)(sin^2x)

h is small quantity neglecting that

-[sin(2x)]/sin^4x

sin2x=2sinxcosx

-[2cosx]/sin^3x

-[2cotx.cosec^2x]

hope it helps u

thank u, that will help me in board never tried first principle

good luck :)