If y - x2 - 1x2 - 1-x2 - - then dydx-

The correct option is B 2xy2y x^2 y = x^2 + 1x^2 + 1x^2 + ... ∞ y = x^2 + 1y y^2 = x^2 y + 1 2y dydx = 2xy + x^2 dydx dydx = 2xy2 ...

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

Implicit Differentiation

If y = x2 +...

Question

If $y = x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + \dots \dots \infty}},$ then $\frac{d y}{d x} =$

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

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

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

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

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

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

\frac{x y^{2}}{y^{2} + 1}

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

Open in App

Solution

Suggest Corrections

Similar questions

If $\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)$ , then show that $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}} . d x}$

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

\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

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

\frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}

x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} = 1

\frac{d y}{d x} = - \sqrt{\frac{{1 - y}^{2}}{{1 - x}^{2}} .}

y = 1 + \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{3}} + . . . . . \infty

| x | > 1

\frac{d y}{d x} =

Join BYJU'S Learning Program