Derivative of Cosecant

We shall prove the formula for the derivative of the cosecant function by using definition or the first principle method.

Let us suppose that the function is of the form

y = f\left( x \right) = \csc x

First we take the increment or small change in the function:

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

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

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

Taking the LCM (Least Common Factor), we get

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

Using the formula from trigonometry, we have

\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)

Using this formula in equation (i), we get

\begin{gathered}\Delta y = - \frac{{\left[ {2\cos \left\{ {\frac{{\left( {x + \Delta x} \right) + x}}{2}} \right\}\sin \left\{ {\frac{{\left( {x + \Delta x} \right) - x}}{2}} \right\}} \right]}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \Delta y = - \frac{{2\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\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{{2\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\Delta x\sin \left( {x + \Delta x} \right)\sin x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = - \frac{{\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}}}{{\sin \left( {x + \Delta x} \right)\sin x}} \times \frac{{\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}

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

Example: Find the derivative of

y = f\left( x \right) = \csc 5x

We have the given function as

y = \csc 5x

Differentiating with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}\csc 5x

Using the rule, \frac{d}{{dx}}\csc x = - \csc x\cot x, we get

\begin{gathered}\frac{{dy}}{{dx}} = - \csc 5x\cot 5x\frac{d}{{dx}}5x \\ \frac{{dy}}{{dx}} = - \csc 5x\cot 5x\left( 5 \right) \\ \frac{{dy}}{{dx}} = - 5\csc 5x\cot 5x \\ \end{gathered}