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

Solution:

Let y = cosec x

⇒ y = 1/sinx

We have d/dx (u/v) = {v(du/dx) - u(dv/dx)} / v²

Here, u = 1 and v = sinx

d/dx(1 / sinx) = {[(sin x ). d/dx(1)] - [(1). d/dx(sin x)]} / (sin x)²

= {[sin x . (0)] - [(1) . (cos x)]} / sin² x

= (-cosx) / sin² x

= - (1 / sin x) . (cos x / sin x)

= - cosec x . cot x

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

Summary:

d/dx (csc x) = -csc x cot x holds true.