Derivative of Cosecant Function/Proof 1

Theorem

$\map {\dfrac \d {\d x} } {\csc x} = -\csc x \cot x$

where $\sin x \ne 0$.

Proof

From the definition of the cosecant function:

$\csc x = \dfrac 1 {\sin x}$

From Derivative of Sine Function:

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$

Then:

\(\ds \map {\dfrac \d {\d x} } {\csc x}\)

\(=\)

\(\ds \cos x \frac {-1} {\sin^2 x}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \frac {-1} {\sin x} \frac {\cos x} {\sin x}\)

\(\ds \)

\(=\)

\(\ds -\csc x \cot x\)

Definition of Real Cosecant Function and Definition of Real Cotangent Function

This is valid only when $\sin x \ne 0$.

$\blacksquare$