Question

We use the derivative according to the given problem .The derivative of function $f(x) = \dfrac{{{x^2}}}{{[1 - si{n^2}x]}}$ is
1. \[Even{\text{ }}function\]
2. \[Odd{\text{ }}function\]
3. \[Not{\text{ }}define\]
4. \[Increasing{\text{ }}Function\]

Answer 1

Hint: We have to derivative the function $f(x) = \dfrac{{{x^2}}}{{[1 - si{n^2}x]}}$ and state the type of function formed after differentiating the function \[f\left( x \right)\] . We solve this by firstly differentiating the function \[f\left( x \right)\] with respect to using chain rule and various basic derivative formulas of trigonometric functions and derivatives of ${x^n}$ . Then in the function \[f'\left( x \right)\] we replace \[\;x\] by \[\left( { - x} \right)\] and then state the type of the function .

Complete step-by-step solution:
Given : $f(x) = \dfrac{{{x^2}}}{{[1 - si{n^2}x]}}$
Let \[f\left( x \right){\text{ }} = {\text{ }}y\]
So ,
$y = f(x) = \dfrac{{{x^2}}}{{[1 - si{n^2}x]}}$
We know that , $si{n^2}x + co{s^2}x = 1$
Using this formula , we get
$co{s^2}x = 1 - si{n^2}x$
Putting this in \[y\] , we get
$y = \dfrac{{{x^2}}}{{co{s^2}x}}$
Also we know that \[cos{\text{ }}x{\text{ }} = \dfrac{1}{{{\text{ }}sec{\text{ }}x}}\]
Hence , the function becomes
$y = {x^2} \times se{c^2}x$
Now we have to derivative of \[y\] with respect to
Using product rule of differentiation , ( derivative of ${x^n} = n \times {x^{(n - 1)}}$ ) and ( derivative of \[sec{\text{ }}x{\text{ }} = {\text{ }}sec{\text{ }}x{\text{ }} \times {\text{ }}tan{\text{ }}x\] ) , we get
$\dfrac{{dy}}{{dx}} = 2xse{c^2}x + 2\sec x \times \tan x \times {x^2}$
Taking elements common , we get
\[\dfrac{{dy}}{{dx}} = {\text{ }}2x{\text{ }}sec{\text{ }}x{\text{ }}\left[ {{\text{ }}sec{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x{\text{ }} \times {\text{ }}x{\text{ }}} \right]\]
The derivative of function \[f\left( x \right){\text{ }} = {\text{ }}f'\left( x \right){\text{ }} = \dfrac{{dy}}{{dx}}\]
So ,
\[f'\left( x \right){\text{ }} = {\text{ }}2x{\text{ }}sec{\text{ }}x{\text{ }}\left[ {{\text{ }}sec{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x{\text{ }} \times {\text{ }}x{\text{ }}} \right]\]
Now , we will replace \[\;x\] by \[ - x\] in \[f'\left( x \right)\] to check whether the function is an even or an odd function
Replacing \[x\] by \[ - x\] , we get
\[f'\left( { - x} \right){\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}\left( { - x{\text{ }}} \right){\text{ }}sec{\text{ }}\left( { - x} \right){\text{ }} \times {\text{ }}\left[ {{\text{ }}sec{\text{ }}\left( { - x} \right){\text{ }} + {\text{ }}tan{\text{ }}\left( { - x} \right){\text{ }} \times {\text{ }}\left( { - x} \right){\text{ }}} \right]\]
We also know that , \[tan{\text{ }}\left( { - x} \right){\text{ }} = {\text{ }} - {\text{ }}tan{\text{ }}x\] and \[sec{\text{ }}\left( { - x} \right){\text{ }} = {\text{ }}sec{\text{ }}x\]
Using this , we get
\[f'\left( { - x} \right){\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}\left( { - x{\text{ }}} \right){\text{ }}sec{\text{ }}x{\text{ }} \times {\text{ }}\left[ {{\text{ }}sec{\text{ }}x{\text{ }} + {\text{ }}\left( { - {\text{ }}tan{\text{ }}x{\text{ }}} \right){\text{ }} \times {\text{ }}\left( { - {\text{ }}x} \right){\text{ }}} \right]\]
Also we know that product of two negative terms gives a positive result
\[f'\left( { - x} \right){\text{ }} = {\text{ }} - {\text{ }}2{\text{ }}x{\text{ }} \times {\text{ }}sec{\text{ }}x{\text{ }}\left[ {{\text{ }}sec{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x{\text{ }} \times {\text{ }}x{\text{ }}} \right]\]
Now , we can conclude that
\[f'\left( { - x} \right){\text{ }} = {\text{ }} - {\text{ }}f\left( x \right)\] we also know that if we get this relation then the function is an odd function .
Hence , the derivative of function \[f\left( x \right)\] is an odd function
Thus , the correct answer is \[\left( 2 \right)\].

Note: If after replacing \[\;x\] by \[ - x\] in \[f'\left( x \right)\] and we would get the result that \[f\left( { - x} \right){\text{ }} = {\text{ }}f\left( x \right)\] , then we would have stated that function is an even function .
We differentiated \[y\] with respect to to find \[\dfrac{{dy}}{{dx}}\] . We know the differentiation of trigonometric function :
\[\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x\]
\[\dfrac{{d\left[ {sin{\text{ }}x} \right]}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x\]
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = se{c^2}x$
Derivative of product of two function is given by the following product rule :
\[\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]