Prove that d /dx (csc x) = -csc x cot x.

Solution:

Given function cosec x

d(cscx)dx = d(1/sinx)dx

= lim_h→0 1/sin(x+h)-1/sinx / (x+h)-x

= lim_h→0 sinx - sin(x + h)/sinxsin(x + h) /h

= lim_h→0 sinx - sin(x + h) / h sinx sin(x + h)

= limh→0 -(sin(x + h) - sinx)/h sinx sin(x + h)

= lim_h→0 - (sin(x + h) - sinx)h × 1/lim_h→0 sinx sin(x + h)

= -cosx × 1/sin²x

= -cosx/sinx × 1/sinx

= -cotx cscx

Therefore, we have proved the derivative of cosec x to be -cot x cosec x using the first principle of differentiation.

Prove that d /dx (csc x) = -csc x cot x.

Summary:

Hence, it is proved that d /dx (csc x) = -csc x cot x.