Differentiate y-x2- 1 x2- 1 x2- 1 x2-

The correct option is A dydx=2xy2y x^2 y= x ^ 2 #10;+ 1 x ^ 2 + 1 x ^ 2 + 1 x ^ 2 #10;+ 1 ... #160; #160; #160; #10; #10; y= ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Theorems on Equal Ratios

Differentiate...

Question

Differentiate $y = x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + . . . . . .}}}$

\frac{d y}{d x} = \frac{y}{2 y - x}

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

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

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

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

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

None of these

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

Open in App

Solution

Suggest Corrections

Similar questions

If y = $\sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + . . . \infty}}}}$ , then $\frac{d y}{d x}$ is equal to

\frac{d y}{d x}

^{2}

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

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

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

\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0

\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0

Join BYJU'S Learning Program