If x^2+y^2=t- 1t ,x^4+y^4=t^2+ 1t^2 then dy dx=

Question

Accepted Answer

Correct answer: D. (d) Given x^2+y^2=t- 1t and x^4+y^4=t^2+ 1t^2 ⇒x^4+y^4=x^4+y^4+2x^2y^2+2 ⇒x^2y^2+1=0 Differentiate it w.r.t.x. 2xy^2+x^22y dy dx1=0 ⇒ dy dx= -2xy^22x^2y= -yx

$If x^{2} + y^{2} = t - \frac{1}{t}, x^{4} + y^{4} = t^{2} + \frac{1}{t ^{2}} then \frac{dy}{dx} =$

Solution

More Sets, Relations & Functions PYQs