Question

If \[x = \dfrac{{\left( {1 - {t^2}} \right)}}{{\left( {1 + {t^2}} \right)}}\] and \[y = \dfrac{{2t}}{{\left( {1 + {t^2}} \right)}}\], then find the value of \[\dfrac{{dy}}{{dx}}\].
A. \[\dfrac{y}{x}\]
B. \[\dfrac{x}{y}\]
C. \[ - \dfrac{x}{y}\]
D. \[ - \dfrac{y}{x}\]

Answer 1

Hint: Differentiate the given value of \[x\]and \[y\] with respect to \[t\] to get \[\dfrac{{dx}}{{dt}}\] and \[\dfrac{{dy}}{{dt}}\] respectively. Divide the obtained value of \[\dfrac{{dy}}{{dt}}\] by \[\dfrac{{dx}}{{dt}}\] to get the resultant value of \[\dfrac{{dy}}{{dx}}\].

Formula Used: \[\cos \theta = \dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]
\[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]

Complete step by step solution: Substitute \[\tan \theta \] for \[t\] in the \[x = \dfrac{{\left( {1 - {t^2}} \right)}}{{\left( {1 + {t^2}} \right)}}\] the given value of \[x\].
\[x = \dfrac{{\left( {1 - {{\left( {\tan \theta } \right)}^2}} \right)}}{{\left( {1 + {{\left( {\tan \theta } \right)}^2}} \right)}}\]
\[x = \dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]
Use trigonometric identity, \[\cos \theta = \dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{\left( {1 + {{\tan }^2}\theta } \right)}}\] in the resultant equation.
\[x = \cos 2\theta \] …(1)
Differentiate the obtained value of \[x\] with respect to \[t\].
\[\dfrac{{dx}}{{d\theta }} = \dfrac{{d\left( {\cos 2\theta } \right)}}{{d\theta }}\]
The derivative of \[y = \cos \left( {ax} \right)\] with respect to \[x\] is \[ - a\sin \left( {ax} \right)\], use this to differentiate the resultant equation.
\[\dfrac{{dx}}{{d\theta }} = 2\sin 2\theta \] … (2)
Substitute \[\tan \theta \] for \[t\] in the \[y = \dfrac{{2t}}{{\left( {1 + {t^2}} \right)}}\] the given value of \[y\].
\[y = \dfrac{{2\left( {\tan \theta } \right)}}{{\left( {1 + {{\left( {\tan \theta } \right)}^2}} \right)}}\]
\[y = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\]
Use trigonometric identity, \[\sin 2\theta = \dfrac{{2\tan \theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}\] in the resultant equation.
\[y = \sin 2\theta \] …(3)
Differentiate the obtained value of \[y\] with respect to \[t\].
\[\dfrac{{dy}}{{d\theta }} = \dfrac{{d\left( {\sin 2\theta } \right)}}{{d\theta }}\]
The derivative of \[y = \sin \left( {ax} \right)\] with respect to \[x\] is \[a\cos \left( {ax} \right)\], use this to differentiate the resultant equation.
\[\dfrac{{dy}}{{d\theta }} = 2\cos 2\theta \] … (4)
Divide the value of \[\dfrac{{dy}}{{d\theta }}\] from equation 4 by \[\dfrac{{dx}}{{d\theta }}\] from equation 2, to get the value of \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}}\]
\[\dfrac{{dy}}{{dx}} = - \dfrac{{2\cos 2\theta }}{{2{\mathop{\rm Sin}\nolimits} 2\theta }}\]
\[\dfrac{{dy}}{{dx}} = - \dfrac{{\cos 2\theta }}{{{\mathop{\rm Sin}\nolimits} 2\theta }}\]
Substitute the value of \[\cos 2\theta \] and \[\sin 2\theta \] from equation (1) and equation (3).
\[\dfrac{{dy}}{{dx}} = - \dfrac{x}{y}\]
Hence, the obtained value of \[\dfrac{{dy}}{{dx}}\] is \[ - \dfrac{x}{y}\]

Option ‘C’ is correct

Note: The obtained value of \[\dfrac{{dy}}{{dx}}\] will not be variable \[x\]. So, Replace the variable \[\theta \] in the form of \[x\] and \[y\].