Question

If \[y = f\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\] and ${f'}(x) = {\sin ^2}x$ , then $\dfrac{{dy}}{{dx}} = $
(1) $\left\{ {\dfrac{{\left( {6{x^2} - 2x + 2} \right)}}{{{{({x^2} + 1)}^2}}}} \right\}\sin \left( {\dfrac{{{{(2x - 1)}^2}}}{{({x^2} + 1)}}} \right)$
(2) $\left\{ {\dfrac{{\left( {6{x^2} - 2x + 2} \right)}}{{{{({x^2} + 1)}^2}}}} \right\}{\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
(3) $\left\{ {\dfrac{{\left( { - 2{x^2} + 2x + 2} \right)}}{{{{({x^2} + 1)}^2}}}} \right\}{\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
(4) $\left\{ {\dfrac{{\left( { - 2{x^2} + 2x + 2} \right)}}{{{{({x^2} + 1)}^2}}}} \right\}\sin {\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)^2}$

Answer 1

Hint: If a function that $ = f(h(x))$ then we get the $\dfrac{{dy}}{{dx}} = f'(h(x)) \times h'(x)$ , where $\dfrac{{dy}}{{dx}}$ is the differentiation of $y$ with respect to $x$ . The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument . Differentiation is a tool of calculus . We differentiate the given function and get the required result .

Complete step by step answer:
From the given data \[y = f\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\] ………………………………….(1)
and ${f'}(x) = {\sin ^2}x$……………………………….(2)
Differentiate the above equation (1) with respect to $x$ , we get
$\dfrac{{dy}}{{dx}} = f'\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right).\dfrac{d}{{dx}}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
From the above equation (2) , we get
${f'}(x) = {\sin ^2}x$
\[ \Rightarrow f'(x) = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\]
Now we combining above two equation and we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\dfrac{d}{{dx}}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
Now we differentiate the remaining function by using the formula of $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right)$ , we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\dfrac{{\dfrac{d}{{dx}}(2x - 1) \times ({x^2} + 1) - (2x - 1) \times \dfrac{d}{{dx}}({x^2} + 1)}}{{{{({x^2} + 1)}^2}}}$
We find the differentiation $\dfrac{d}{{dx}}(2x - 1)$
$ = 2$
And the differentiation of $\dfrac{d}{{dx}}({x^2} + 1)$
$ = 2x$
We use this in the above equation and calculate
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\dfrac{{2{x^2} + 2 - 2x(2x - 1)}}{{{{({x^2} + 1)}^2}}}\]
$ \Rightarrow \dfrac{{dy}}{{dx}} = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\dfrac{{2{x^2} + 2 - 4{x^2} + 2x}}{{{{({x^2} + 1)}^2}}}$
Now we arrange this according to the options and get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{( - 2{x^2} + 2x + 2)}}{{{{({x^2} + 1)}^2}}}{\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
Therefore, option (3) is correct.

Note:
In this problem we find the derivative of the composition function and apply the normal derivative rule and then we get the required answer . The differentiation formula of $\left( {\dfrac{u}{v}} \right)$ is $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v \times \dfrac{{du}}{{dx}} - u \times \dfrac{{dv}}{{dx}}}}{{{v^2}}}$ . We can find the solution by another way , first we find the $\dfrac{{dh}}{{dx}}$ and collect the given equations and combine them and get the required answer .
$\dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
$ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{{2{x^2} + 2 - 2x(2x - 1)}}{{{{({x^2} + 1)}^2}}}$
\[ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{{( - 2{x^2} + 2x + 2)}}{{{{({x^2} + 1)}^2}}}\]
From given data \[y = f\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\]
Differentiating above equation and get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)\dfrac{{dh}}{{dx}}\]
Combining above two equations and we get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{( - 2{x^2} + 2x + 2)}}{{{{({x^2} + 1)}^2}}}{\sin ^2}\left( {\dfrac{{(2x - 1)}}{{({x^2} + 1)}}} \right)$
This is the required answer.