Question
If $$y = f\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)$$ and $$f'\left( x \right) = \sin \,{x^2},$$ then $$\frac{{dy}}{{dx}}$$ is :
A.
$$\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}$$
B.
$$\frac{{2\left( {1 + x - {x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}$$
C.
$$\frac{{2\left( {2x - 1} \right)}}{{{x^2} + 1}}\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}$$
D.
none of these
Answer :
$$\frac{{2\left( {1 + x - {x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}$$
Solution :
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{df\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{dx}} = \frac{{df\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{d\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}.\frac{{d\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)}}{{dx}} \cr
& = \sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2}.\frac{{2\left( {{x^2} + 1} \right) - \left( {2x - 1} \right)2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& = \frac{{2\left( {1 + x - {x^2}} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\sin {\left( {\frac{{2x - 1}}{{{x^2} + 1}}} \right)^2} \cr} $$