What is the Derivative of csc(x)?

derivative of csc(x)

What is the derivative of csc(x)?

We will see that the derivative of

\csc(x)

-\csc(x)\cot(x)

.

Proof. Let

F(x) = \csc(x) = \frac{1}{\sin(x)}

f(u) = \frac{1}{u}

and

g(x) = \sin(x)

such that

F(x) = f(g(x))

. Then we will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We have seen earlier that here that

g'(x) = \frac{d}{dx} \sin(x) = \cos(x)

and

f'(u) = \frac{-1}{u^2}

. So we get

\begin{align*} f'(g(x)) = \frac{-1}{g(x)^2} = \frac{-1}{\sin^2(x)} \quad \text{and} \quad g'(x) = \cos(x). \end{align*}

Wrapping everything together, we get

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{-1}{\sin^2(x)}\cdot \cos(x) \\ &= -\frac{\cos(x)}{\sin^2(x)} \\ &= -\frac{\cos(x)}{\sin(x)}\frac{1}{\sin(x)} \\ &= -\csc(x) \frac{1}{\cos(x)} \\ &= -\csc(x)\cot(x) \end{align*}

In conclusion, the derivative of

\csc(x)

-\csc(x)\cot(x)