• Algebra
• Trigonometry
• Geometry
• Calculus

Proof of Derivative of cotx formula

The derivative of cot function with respect to a variable is equal to the negative of square of cosecant function. If $x$ is taken as a variable and represents an angle of a right triangle, then the co-tangent function is written as $\cot{x}$ in mathematics. The differentiation of the $\cot{x}$ function with respect to $x$ is equal to $-\csc^2{x}$ or $-\operatorname{cosec}^2{x}$, and it can be proved from first principle in differential calculus.

Differentiation of function in Limit form

On the basis of definition of the derivative, the derivative of a function in terms of $x$ can be written in the following limits form.

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

Here, if $f{(x)} = \cot{x}$, then $f{(x+h)} = \cot{(x+h)}$. Now, let’s find the proof of the differentiation of $\cot{x}$ function with respect to $x$ by the first principle.

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

Simplify the entire function

The difference of the cot functions in the numerator can be simplified by the quotient rule of cos and sin functions.

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

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

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

Actually, the trigonometric expression in the numerator is actually the expansion of angle difference identity of the sin function.

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

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

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

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

According to Even/ Odd trigonometric identity of sin function, the sine of negative angle is equal to negative of sin of angle.

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

Now, split the limit of the function as the limit of product of two trigonometric functions.

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

Try product rule of limits for evaluating limit of product of the two functions.

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

Evaluate Limits of trigonometric functions

Now, use limit of sinx/x as x approaches 0 formula, and the limit of the first trigonometric function is equal to one as $h$ approaches $0$.

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

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

Now, the limit of the trigonometric function can be evaluated by the direct substitution method.

$=\,\,\, \dfrac{-1}{\sin{(x+0)}\sin{x}}$

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

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

The reciprocal of the sin function is equal to cosecant according to the reciprocal identity of sine function.

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

Therefore, it is derived by differentiation from first principle that the differentiation of cot function with respect to a variable is equal to negative square of cosecant function.