MCQ
If $x = y\sqrt {1 - {y^2},} $ then ${{dy} \over {dx}} = $
- A$0$
- B$x$
- ✓${{\sqrt {1 - {y^2}} } \over {1 - 2{y^2}}}$
- D${{\sqrt {1 - {y^2}} } \over {1 + 2{y^2}}}$
✓
Answer
Correct option: C.
${{\sqrt {1 - {y^2}} } \over {1 - 2{y^2}}}$
c
(c) $x = y\sqrt {1 - {y^2}} $
(c) $x = y\sqrt {1 - {y^2}} $
Differentiate with respect to $x,$
$1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} + y.\frac{1}{{2\sqrt {1 - {y^2}} }}\,.\,( - 2y)\,.\,\frac{{dy}}{{dx}}$
==> $1 = \frac{{dy}}{{dx}}\sqrt {1 - {y^2}} - \frac{{{y^2}}}{{\sqrt {1 - {y^2}} }}\,.\,\frac{{dy}}{{dx}}$
==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - {y^2} - {y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$
==> $1 = \frac{{dy}}{{dx}}\left[ {\frac{{1 - 2{y^2}}}{{\sqrt {1 - {y^2}} }}} \right]$
$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}$.
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