Question

How do you differentiate \[f\left( x \right)=\dfrac{1+{{\sin }^{2}}x}{1-\sin 2x}\] using the quotient rule?

Answer 1

Hint: From the question given, we have been asked to differentiate \[f\left( x \right)=\dfrac{1+{{\sin }^{2}}x}{1-\sin 2x}\] using the quotient rule. We can solve the above given question by knowing the formula for the quotient rule. Formula for the quotient rule is shown below:
\[{f}'\left( x \right)=\dfrac{h\left( x \right){g}'\left( x \right)-{h}'\left( x \right)g\left( x \right)}{{{\left( h\left( x \right) \right)}^{2}}}\] We can clearly observe that the given function is in the form of \[f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}\].

Complete step by step answer:
For answering this question we will be using the quotient rule given as \[{f}'\left( x \right)=\dfrac{h\left( x \right){g}'\left( x \right)-{h}'\left( x \right)g\left( x \right)}{{{\left( h\left( x \right) \right)}^{2}}}\] for a function in the form of $f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}$ .
Now, we have to find out the all terms needed to substitute in the quotient rule formula.
First of all, from the question, we have been given that, \[g\left( x \right)=1+{{\sin }^{2}}x\]
Also, we have to find the derivative for it.
Therefore,
\[\Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}{{\left( \sin x \right)}^{2}}\]
\[\Rightarrow {g}'\left( x \right)=2\cos x\sin x\]
Also, from the question given we have been given that \[h\left( x \right)=1-\sin 2x\]
Also, we have to find the derivative for it.
Therefore,
\[\Rightarrow {h}'\left( x \right)=\dfrac{d}{dx}\left( 1 \right)-\dfrac{d}{dx}\left( sin2x \right)\]
\[\Rightarrow {h}'\left( x \right)=-2\cos 2x\]
Therefore, we got all the terms which are needed to substitute in the quotient rule formula.
Now, substitute all the terms we got in the quotient rule formula. By substituting the all the terms we got in the quotient rule formula, we get
\[{f}'\left( x \right)=\dfrac{h\left( x \right){g}'\left( x \right)-{h}'\left( x \right)g\left( x \right)}{{{\left( h\left( x \right) \right)}^{2}}}\]
\[\Rightarrow {f}'\left( x \right)=\dfrac{2\cos x\sin x\left( 1-\sin 2x \right)+2\cos 2x\left( 1+{{\sin }^{2}}x \right)}{{{\left( 1-\sin 2x \right)}^{2}}}\]
\[\Rightarrow {f}'\left( x \right)=\dfrac{2\cos x\sin x}{\left( 1-\sin 2x \right)}+\dfrac{2\cos 2x\left( 1+{{\sin }^{2}}x \right)}{{{\left( 1-\sin 2x \right)}^{2}}}\]
Hence, derivative for the given function was found by using the quotient rule formula.

Note: We should be well known about the differentiation concept. Also, we should do the calculation very carefully especially while finding the derivatives of the function. We should be well known about the basic formulae of differentiation. In this question, we have used the quotient rule formula. So, we should have some basic knowledge on the formulae of differentiation. Also, we should be very careful while finding the derivatives of the functions. Similarly the product rule for a function in the form of $f\left( x \right)=g\left( x \right)\times h\left( x \right)$ is given as ${{f}^{'}}\left( x \right)={{g}^{'}}\left( x \right)\times h\left( x \right)+{{h}^{'}}\left( x \right)\times g\left( x \right)$