MCQ
If $\ln \,(x + y) = 2xy,$ then $y'(0) =$
- ✓$1$
- B$-1$
- C$2$
- D$0$
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 $y'(0) = \frac{{1 - 2}}{{ - 1}} = 1$.
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