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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
If ${x^2} + {y^2} = t - \frac{1}{t},{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$ then $\frac{{dy}}{{dx}}$ equals
  • A
    $1/x{y^3}$
  • ✓
    $1/{x^3}y$
  • C
    $ - 1/{x^3}y$
  • D
    $ - 1/x{y^3}$
✓

Answer

Correct option: B.
$1/{x^3}y$
b
(b) ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$
${({x^2})^2} + {({y^2})^2} = {t^2} + \frac{1}{{{t^2}}}$
==> ${({x^2} + {y^2})^2} - 2{x^2}{y^2} = {t^2} + \frac{1}{{{t^2}}}$
==> ${\left( {t - \frac{1}{t}} \right)^2} - 2{x^2}{y^2} = {t^2} + \frac{1}{{{t^2}}}$ ==> ${x^2}{y^2} = - 1$ ==>${y^2} = \frac{{ - 1}}{{{x^2}}}$
Differentiation w.r.t. $x$, we get $2y\frac{{dy}}{{dx}} = \frac{2}{{{x^3}}} \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{{x^3}y}}$.

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