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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
${d \over {dx}}{\tan ^{ - 1}}{{4\sqrt x } \over {1 - 4x}} = $
  • A
    ${1 \over {\sqrt x (1 + 4x)}}$
  • ✓
    ${2 \over {\sqrt x (1 + 4x)}}$
  • C
    ${4 \over {\sqrt x (1 + 4x)}}$
  • D
    None of these
✓

Answer

Correct option: B.
${2 \over {\sqrt x (1 + 4x)}}$
b
(b) $\frac{d}{{dx}}{\tan ^{ - 1}}\frac{{4\sqrt x }}{{1 - 4x}}$

$ = \frac{1}{{1 + {{\left( {\frac{{4\sqrt x }}{{1 - 4x}}} \right)}^2}}}.\left[ {\frac{{(1 - 4x)4\left( {\frac{1}{{2\sqrt x }}} \right) - 4\sqrt x ( - 4)}}{{{{(1 - 4x)}^2}}}} \right]$

$ = \frac{{2(1 + 4x)}}{{\sqrt x {{(1 + 4x)}^2}}} = \frac{2}{{\sqrt x (1 + 4x)}}$.

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