You visited us 1 times! Enjoying our articles? Unlock Full Access!

Change of Variables

lf 2tanh-1 ...

Question

lf $2 t a n h^{-} 1 =$ logy,then the value of $y$ in terms of $x$ is

2 x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 x}{1 - x^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

X^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{1 + x}{1 - x})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $(\frac{1 + x}{1 - x})$
$tan h^{- 1} x = \frac{1}{2} ln (\frac{1 + x}{1 - x})$ [formula]
$2 tan h^{- 1} (x) = ln (\frac{1 + x}{1 - x}) = ln y$ [Given]
$y = \frac{1 + x}{1 - y}$

Suggest Corrections

Similar questions

y = \frac{x^{2}}{2} + \frac{1}{2} x \sqrt{x^{2} + 1} + ln \sqrt{x + \sqrt{x^{2} + 1}}

x y^{'} + log y^{'}

y = \frac{1}{2}

(\frac{1}{2_{X} \sqrt{1 + x^{2}}})

x =

2 tan h^{- 1} x = log y

y

x

x, y, z

x + log (x + \sqrt{x^{2} + 1}) = y

y + log (y + \sqrt{y^{2} + 1}) = z

z + log (z + \sqrt{z^{2} + 1}) = x

{(1 - x)}^{2} + {(1 - y)}^{2} + {(1 - z)}^{2}

If $y = [\frac{x^{2} + 1}{x + 1}], t h e n \frac{d y}{d x}$ =?

Join BYJU'S Learning Program