Derivative of Cosecant

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


Differentiation 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}

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