Math Doubts

Derivative of cotx formula Proof

$x$ is an angle of a right angled triangle and then cotangent function is $\cot{x}$. The differentiation or derivative of $\cot{x}$ function with respect to $x$ is written as $\dfrac{d}{dx} \cot{x}$ in differential calculus.

Express Differentiation of function in Limit form

The derivative (or) differentiation of $cot{x}$ function is derived in differential calculus by expressing derivative of $\cot{x}$ function in limit form.

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

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 limit form.

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

Transform tan functions in terms of cos and sin functions

According to quotient identity of cot function with sin and cos functions, each cot function can be written as the quotient of cosine by sine function.

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

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

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

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

Use Compound angle trigonometric identity

The trigonometric expression in numerator represents sine of compound angle identity. So, the entire expression can be minimized by writing it as sin of compound angle.

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

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

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

Split the expression as multiplying factors

Part of the trigonometric expression is same as the limit sinx/x as x approaches 0 formula. So, divide the trigonometric expression as two multiplying factors for taking first step in simplifying it mathematically.

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

Apply Limit Product Rule

The limit value belongs to both multiplying factors. So, it can be applied to both functions to find value of them.

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

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

Simplifying the Expression

The first multiplying factor represents lim sinx/x as x approaches 0 formula. So, its value is equal to 1 and substitute h is equal to 0 to obtain the value of the second multiplying factor. The entire mathematical expression is derivative of cotx function with respect to x.

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

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

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

As per reciprocal identity of sin function with cosecant function, the reciprocal of sin function can be written as cosecant function.

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

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

It is also written as follows.

$\implies \dfrac{d}{dx} \cot{x} \,=\, -\operatorname{cosec}^2{x}$

Therefore, it is proved that the differentiation (or) derivative of $\cot{x}$ with respect to $x$ is negative of square of cosecant of angle $x$.



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