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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $y = \sqrt {{{1 + x} \over {1 - x}}} ,$ then ${{dy} \over {dx}} = $
  • A
    ${2 \over {{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$
  • ✓
    ${1 \over {{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$
  • C
    ${1 \over {2{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$
  • D
    ${1 \over {{{(1 + x)}^{3/2}}{{(1 - x)}^{1/2}}}}$
✓

Answer

Correct option: B.
${1 \over {{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$
b
(b) $y = \sqrt {\frac{{1 + x}}{{1 - x}}} $

==>$y = \sqrt {\frac{{(1 + x)(1 + x)}}{{(1 - x)(1 + x)}}} = \sqrt {\frac{{{{(1 + x)}^2}}}{{1 - {x^2}}}} $

Differentiating with respect to $x,$ we get

$\frac{{dy}}{{dx}} = \frac{{{{(1 - x)}^{1/2}}\frac{d}{{dx}}{{(1 + x)}^{1/2}} - {{(1 + x)}^{1/2}}\frac{d}{{dx}}{{(1 - x)}^{1/2}}}}{{(1 - x)}}$

$ = \frac{{(1 - x) + (1 + x)}}{{2{{(1 - x)}^{3/2}}{{(1 + x)}^{1/2}}}}$

$x = \frac{2}{3}y$.

Trick : $\log y = \frac{1}{2}\log (1 + x) - \frac{1}{2}\log (1 - x)$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2(1 + x)}} + \frac{1}{{2(1 - x)}}$

==> $\frac{{dy}}{{dx}} = \frac{1}{{(1 + x)(1 - x)}} \times \sqrt {\frac{{1 + x}}{{1 - x}}} $

$ = \frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$.

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