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

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 :
$\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$