MCQ
If $\ln \,(x + y) = 2xy,$ then $y'(0) =$
  • $1$
  • B
    $-1$
  • C
    $2$
  • D
    $0$

Answer

Correct option: A.
$1$
a
(a) $\ln (x + y) = 2xy$

Differentiate both sides w.r.t  $x,$

$\left( {\frac{1}{{x + y}}} \right)\,\left( {1 + \frac{{dy}}{{dx}}} \right) = 2\,\left( {x\frac{{dy}}{{dx}} + y} \right)$

==> $\frac{{dy}}{{dx}} = \frac{{1 - 2xy - 2{y^2}}}{{2{x^2} + 2xy - 1}}$

As at $x = 0,y = 1$, (From $\ln (x + y) = 2xy$)

Hence $y'(0) = \frac{{1 - 2}}{{ - 1}} = 1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free