# Derivative of cot x with respect to x

## Formula

$\Large \dfrac{d}{dx} \, \cot x = -csc^2 x$

### Proof

$x$ is an angle of a right angle triangle then the cotangent of angle $x$ is written as $\cot x$ trigonometrically. The derivative of $\cot x$ with respect to same angle is written as $\dfrac{d}{dx} \cot x$ in differentiation calculus.

The derivative of $\cot x$ with respect to $x$ can be derived in mathematics by the principle rule of differentiation in limits form.

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

###### Step: 1

Take $f(x) = \cot x$, then $f(x+h) = \cot(x+h)$ and substitute them in the fundamental principle of differentiation of the function in limits form.

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

###### Step: 2

Write cot of angle as the ratio of cosine of angle to sine of angle as per quotient relation of sine and cosine functions with cotangent.

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

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

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

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{ -[\sin(x+h) \cos x-\cos(x+h) \sin x]}{h\sin(x+h)\sin x}$

###### Step: 3

The trigonometric expression in numerator is the expansion of the sine of difference of two angles.

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

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\require{cancel} \displaystyle \lim_{h \to 0} \dfrac{-\sin(\cancel{x}+h-\cancel{x})}{h\sin(x+h)\sin x}$

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

###### Step: 4

The numerator contains sine of angle $h$ but the same angle is in denominator. So, split the function into two sub functions to apply limit ratio of sine of angle to angle rule when angle tends to zero.

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{-\sin h}{h} \times \dfrac{1}{\sin(x+h)\sin x}$

Apply limit to both multiplying factors.

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\displaystyle \lim_{h \to 0} \dfrac{-\sin h}{h}$ $\times$ $\displaystyle \lim_{h \to 0} \dfrac{1}{\sin(x+h)\sin x}$

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\displaystyle -\lim_{h \to 0} \dfrac{\sin h}{h}$ $\times$ $\displaystyle \lim_{h \to 0} \dfrac{1}{\sin(x+h)\sin x}$

###### Step: 5

According to limit sinx/x rule when x tends to zero, the ratio of sine of angle $h$ to $h$ is $1$ when the value of $h$ approaches $0$. Substitute $h = 0$ in the second multiplying function.

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $-1 \times \dfrac{1}{\sin(x+0)\sin x}$

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $-1 \times \dfrac{1}{\sin x \sin x}$

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $-1 \times \dfrac{1}{\sin^2 x}$

The reciprocal of sine is cosecant as per the reciprocal relation between them.

$\implies \dfrac{d}{dx} \cot x$ $\,=\,$ $\dfrac{-1}{\sin^2 x}$

$\therefore \,\,\,\,\,\, \dfrac{d}{dx} \cot x$ $\,=\,$ $-\csc^2 x$

Therefore, it is proved that the derivative of cot $x$ with respect to $x$ is minus of square of cosecant of angle $x$.

Save (or) Share