Question

How to find the derivative of $ \left( {{\csc }^{2}}x \right) $ ?

Answer 1

Hint: In this problem we have to find derivation of $ \left( {{\csc }^{2}}x \right) $ .In the given expression we can observe the trigonometric ratio i.e., $ \csc x $ .According to the first principle of derivation we will assume the expression $ y=\left( {{\csc }^{2}}x \right) $ take it as equation number one. Then we will increase the expression by adding $ \delta y $ on left side, $ \delta x $ on right side. Now we will consider this as equation number two and subtract both the equations. Now we will use trigonometric identity and simplify the equation. After that we will apply the limit $ \displaystyle \lim_{\delta y\to 0} $ on the left side and $ \displaystyle \lim_{\delta x \to 0} $ on the right side. After applying the limit, we will simplify the obtained equation to get the result.

Complete step by step answer:
Given that, $ {{\csc }^{2}}x$ .
Let us assume $ y={{\csc }^{2}}x.....\left( \text{i} \right) $
Adding $ \delta y $ on left side and $ \delta x $ on right side, then we will get
 $ y+\delta y={{\csc }^{2}}\left( x+\delta x \right).....\left( \text{ii} \right) $
Subtracting equation $ \left( \text{i} \right) $ from the equation $ \left( \text{ii} \right) $ , then we will get
 $ y+\delta y-y={{\csc }^{2}}\left( x+\delta x \right)-{{\csc }^{2}}x $
Simplifying the above equation, then we will get
 $ \delta y={{\csc }^{2}}\left( x+\delta x \right)-{{\csc }^{2}}x $
We know that $ \csc x=\dfrac{1}{\sin x} $ applying this value in the above equation, then we will get
 $ \delta y=\dfrac{1}{{{\sin }^{2}}\left( x+\delta x \right)}-\dfrac{1}{{{\sin }^{2}}x} $
Doing LCM and simplifying the above equation, then we will have
 $ \delta y=\dfrac{{{\sin }^{2}}x-{{\sin }^{2}}\left( x+\delta x \right)}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+\delta x \right)} $
Using the trigonometric formula that is $ \begin{align}
  & {{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right)\sin \left( A-B \right) \\
 & \\
\end{align} $ , then we will get
 $ \delta y=\dfrac{\sin \left( x+x+\delta x \right)\sin \left( x-x-\delta x \right)}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+\delta x \right)} $
We know that $ a+a=2a $ , $ +a-a=0 $ , then we will write the above equation as
 $ \delta y=\dfrac{\sin \left( 2x+\delta x \right)\sin \left( -\delta x \right)}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+\delta x \right)} $
We know that $ \sin \left( -\theta \right)=-\sin \left( \theta \right) $ , then the above equation modified as
 $ \Rightarrow \delta y=\dfrac{-\sin \left( 2x+\delta x \right)\sin \left( \delta x \right)}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+\delta x \right)} $
Now applying limits on both sides, then
 $ \displaystyle \lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}=\displaystyle \lim_{\delta x \to 0}\dfrac{-\sin \left( 2x+\delta x \right)\dfrac{\sin \delta x}{\delta x}}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+\delta x \right)} $
Applying the limit value in the above equation, then we will get
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{-\sin \left( 2x+0 \right)\dfrac{\sin \delta x}{\delta x}}{{{\sin }^{2}}x{{\sin }^{2}}\left( x+0 \right)} $
We know that $ \displaystyle \lim_{\theta \to 0}\dfrac{\sin \theta }{\theta }=1 $ then $ \dfrac{\sin \delta x}{x}=1 $
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{-\sin 2x}{{{\sin }^{2}}x{{\sin }^{2}}x} $
We know that $ \sin 2x=2\sin x\cos x $ ,then
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{-2\sin x\cos x}{{{\sin }^{2}}x{{\sin }^{2}}x} $
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{-2\cos x}{{{\sin }^{2}}x.\sin x} $
We know that $ \dfrac{1}{\sin x}=\csc x $ and $ \dfrac{\cos x}{\sin x}=\cot x $ , then we will get
 $ \dfrac{dy}{dx}=-2\cot x{{\csc }^{2}}x $
$ \therefore $ The final derivation of $ {{\csc }^{2}}x $ is $ -2\cot x{{\csc }^{2}}x $ .

Note:
In the above problem we have used the concept of limits to calculate the derivative of the $ {{\csc }^{2}}x $ . But it is not necessary to use the above procedure to calculate the derivative. We have a simple method to calculate the derivatives.
Given that, $ y={{\csc }^{2}}x $
Differentiating the above equation with respect to $ x $ , then we will get
 $ \dfrac{dy}{dx}={{\left( {{\csc }^{2}}x \right)}^{'}} $
We know that $ {{\left( {{x}^{n}} \right)}^{'}}=n{{x}^{n-1}} $ , then we will get
 $ \dfrac{dy}{dx}=2\left( {{\csc }^{2-1}}x \right){{\left( \csc x \right)}^{'}} $
We know that $ \dfrac{d}{dx}\left( \csc x \right)=-\csc x.\cot x $ , then we will have
 $ \begin{align}
  & \dfrac{dy}{dx}=2\left( \csc x \right)\left( -\csc x.\cot x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=-2{{\csc }^{2}}x\cot x \\
\end{align} $
From both the methods we got the same result.