# Derivative of csc x with respect to x

## Formula

$\large \dfrac{d}{dx} \, \csc x = -\csc x \cot x$

### Proof

Assume, $x$ is an angle of a right angled triangle and the cosecant of angle $x$ is written $\csc x$. The derivative of $\csc x$ with respect to $x$ can be derived mathematically by using the differentiation rule of a function in limit form.

$\dfrac{d}{dx} f(x)$ $=$ $\large \displaystyle \lim_{h \to 0}$ $\dfrac{f(x+h)-f(x)}{h}$

Take $f(x) = \csc x$, then $f(x) = \csc(x+h)$. Now, substitute them in the differentiation rule of the function in limit form.

$\implies$ $\dfrac{d}{dx} \csc x$ $\,=\,$ $\large \displaystyle \lim_{h \to 0}$ $\dfrac{\csc(x+h) -\csc x}{h}$

###### Step: 1

There are no trigonometric identities to simplify the numerator. So, convert each cosecant function in terms of its reciprocal function according to reciprocal relation between sine and cosecant functions. Now, transform the entire function in terms of the sine functions.

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\dfrac{1}{\sin(x+h)} -\dfrac{1}{\sin x}}{h}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\dfrac{\sin x -\sin(x+h)}{\sin(x+h) \sin x}}{h}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\sin x -\sin(x+h)}{h \sin(x+h) \sin x}$

###### Step: 2

Use sum to product transformation identity to express the numerator in product form.

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{2 \cos \Bigg[ \dfrac{x+x+h}{2} \Bigg] \sin \Bigg[ \dfrac{x-(x+h)}{2} \Bigg]}{h \sin(x+h) \sin x}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{2 \cos \Bigg[ \dfrac{2x+h}{2} \Bigg] \sin \Bigg[ \dfrac{x-x-h}{2} \Bigg]}{h \sin(x+h) \sin x}$

$\,=\,$ $\require{cancel} \large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{2 \cos \Bigg[ \dfrac{2x+h}{2} \Bigg] \sin \Bigg[ \dfrac{\cancel{x}-\cancel{x}-h}{2} \Bigg]}{h \sin(x+h) \sin x}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{2 \cos \Bigg[ \dfrac{2x+h}{2} \Bigg] \sin \Bigg[ \dfrac{-h}{2} \Bigg]}{h \sin(x+h) \sin x}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{-2 \cos \Bigg[ \dfrac{2x+h}{2} \Bigg] \sin \Bigg[ \dfrac{h}{2} \Bigg]}{h \sin(x+h) \sin x}$

$\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{-2 \sin \Bigg[ \dfrac{h}{2} \Bigg] \cos \Bigg[ \dfrac{2x+h}{2} \Bigg]}{h \sin(x+h) \sin x}$

###### Step: 3

Now, divide the entire function as two multiplying sub functions to use limit sinx/x rule when x tends to 0.

$\,=\,$ $- \large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{2 \sin \Bigg[ \dfrac{h}{2} \Bigg] }{h}$ $\times$ $\dfrac{\cos \Bigg[ \dfrac{2x+h}{2} \Bigg]}{\sin(x+h) \sin x}$

$\,=\,$ $- \large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\sin \Bigg[ \dfrac{h}{2} \Bigg] }{\dfrac{h}{2}}$ $\times$ $\dfrac{\cos \Bigg[ \dfrac{2x+h}{2} \Bigg]}{\sin(x+h) \sin x}$

$\,=\,$ $- \large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\sin \Bigg[ \dfrac{h}{2} \Bigg] }{\dfrac{h}{2}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}$ $\dfrac{\cos \Bigg[ \dfrac{2x+h}{2} \Bigg]}{\sin(x+h) \sin x}$

###### Step: 4

If $h \,\to\, 0$ then $\dfrac{h}{2} \,\to\, \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \,\to\, 0$. Now, change the limit of first function but no need to change the limit of second multiplying function.

$\,=\,$ $- \large \displaystyle \lim_{\frac{h}{2} \to 0}$ $\dfrac{\sin \Bigg[ \dfrac{h}{2} \Bigg] }{\dfrac{h}{2}}$ $\times$ $\large \displaystyle \lim_{h \to 0}$ $\dfrac{\cos \Bigg[ \dfrac{2x+h}{2} \Bigg]}{\sin(x+h) \sin x}$

###### Step: 5

According to the limit $\dfrac{\sin x}{x}$ rule when $x$ tends to $0$, the value of the quotient of $\sin \Big(\dfrac{h}{2}\Big)$ by $\dfrac{h}{2}$ is one if $\dfrac{h}{2}$ tends to $0$. Similarly, substitute $h$ equals to $0$ in the second multiplying limit function.

$\,=\,$ $-1 \times \dfrac{\cos \Bigg[ \dfrac{2x+0}{2} \Bigg]}{\sin(x+0) \sin x}$

$\,=\,$ $-\dfrac{\cos \Bigg[ \dfrac{2x}{2} \Bigg]}{\sin x \times \sin x}$

$\,=\,$ $\require{cancel} -\dfrac{\cos \Bigg[ \dfrac{\cancel{2}x}{\cancel{2}} \Bigg]}{\sin x \sin x}$

$\,=\,$ $-\dfrac{\cos x }{\sin x \sin x}$

$\,=\,$ $-\dfrac{1}{\sin x} \times \dfrac{\cos x }{\sin x}$

$\,=\,$ $-\csc x \cot x$

$\therefore \,\,\,\,\,\, \dfrac{d}{dx} \, \csc x = -\csc x \cot x$

Therefore, it has proved that the derivative of cosecant of angle $x$ with respect to $x$ is equal to the negative of product of cosecant of $x$ and cotangent of $x$.