MCQ
$x = \frac{{{e^{2y}} - 1}}{{{e^{2y}} + 1}}$ તો $\frac{{dy}}{{dx}} = ......$
- A$1 + {x^2}$
- B$\frac{1}{{{x^2} - 1}}$
- ✓$\frac{1}{{1 - {x^2}}}$
- D${x^2} - 1$
$x=\frac{e^{2y}-1}{e^{2y}+1}\\x,(e^{2y}+1)=e^{2y}-1\\xe^{2y}+x-e^{2y}=-1\\e^{2y}(x-1)=-1-x$
$e^{2y}=\frac{-1-x}{x-1}\Rightarrow2y=log\left(\frac{-x-1}{x-1}\right)$
$2\frac{dy}{dx}=\frac{1}{\frac{-x-1}{x-1}}.\frac{2}{(x-1)^2}$
$2\frac{dy}{dx}=\frac{x-1}{-x-1}\times\frac{2}{(x-1)^2}$
$\frac{dy}{dx}=\frac{1}{(-x-1)(x-1)}=\frac{1}{1-x^2}$
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