Derivative of Cotangent

We shall prove formula for derivative of cotangent function using by definition or first principle method.
Let us suppose that the function of the form y = f\left( x \right) =  \cot x.
First we take the increment or small change in the function.

\begin{gathered} y + \Delta y = \cot \left( {x + \Delta x}  \right) \\ \Delta y = \cot \left( {x + \Delta x} \right)  - y \\ \end{gathered}


Putting the value of function $y = \cot x$ in the above equation, we get

\begin{gathered} \Delta y = \cot \left( {x + \Delta x} \right)  - \cot x \\ \Delta y = \frac{{\cos \left( {x + \Delta x}  \right)}}{{\sin \left( {x + \Delta x} \right)}} - \frac{{\cos x}}{{\sin x}} \\ \Delta y = \frac{{\cos \left( {x + \Delta x}  \right)\sin x - \sin \left( {x + \Delta x} \right)\cos x}}{{\sin \left( {x +  \Delta x} \right)\sin x}} \\ \Delta y = - \frac{{\sin \left( {x + \Delta x} \right)\cos x - \cos \left( {x +  \Delta x} \right)\sin x}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \end{gathered}


Using formula from trigonometry, we have

\sin  \left( {\alpha - \beta } \right) = \sin  \alpha \cos \beta - \cos \alpha \sin  \beta


Using this formula in equation (i), we get

\begin{gathered} \Delta y = - \frac{{\sin \left( {x + \Delta x - x} \right)}}{{\sin \left( {x +  \Delta x} \right)\sin x}} \\ \Delta y = - \frac{{\sin \Delta x}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \end{gathered}


Dividing both sides by \Delta  x, we get

\begin{gathered} \Rightarrow \frac{{\Delta y}}{{\Delta x}}  = - \frac{{\sin \Delta x}}{{\Delta x\sin  \left( {x + \Delta x} \right)\sin x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}}  = - \frac{1}{{\sin \left( {x + \Delta x}  \right)\sin x}} \times \frac{{\sin \Delta x}}{{\Delta x}} \\ \end{gathered}


Taking limit of both sides as \Delta x \to 0, we have

\begin{gathered}\Rightarrow \mathop {\lim }\limits_{\Delta x  \to 0} \frac{{\Delta y}}{{\Delta x}} = -  \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\sin \left( {x + \Delta x}  \right)\sin x}} \times \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin  \Delta x}}{{\Delta x}} \\ \frac{{dy}}{{dx}} = - \frac{1}{{\sin \left( {x + 0} \right)\sin  x}} \times \left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = - \frac{1}{{{{\sin }^2}x}} \\ \Rightarrow \frac{{dy}}{{dx}} = - {\csc ^2}x \\ \Rightarrow \frac{d}{{dx}}\left( {\cot x}  \right) = - {\csc ^2}x \\ \end{gathered}

Example: Find the derivative of

y = f\left( x \right)  = \cot {x^2}


We have the given function as

y =  \tan 2x


Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}}  = \frac{d}{{dx}}\cot {x^2}


Using the rule, $\frac{d}{{dx}}\left( {\cot x} \right) =  - {\csc ^2}x$, we get

\begin{gathered} \frac{{dy}}{{dx}} = - {\csc ^2}{x^2}\frac{d}{{dx}}{x^2} \\ \frac{{dy}}{{dx}} = - {\csc ^2}{x^2} \times 2x \\ \frac{{dy}}{{dx}} = - 2x{\csc ^2}{x^2} \\ \end{gathered}

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