Question

If x^2+y^2=t-1/t and x^4+y^4=t^2+1/t^2 then find x^3ydy/dx

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

8x^2+y^2=t-1/t & x^4+y^4=t^2+1/t^2 now let us evaluate x^4+y^4=(x^2+y^2)^2–2x^2y^2

or,t^2+1/t^2=t^2+1/t^2-2-2(xy)^2

or,t^2-t^2+1/t^2-1/t^2+2=-2(xy)^2

or,2(xy)^2=-2

or,y^2=-1/x^2 (by differentiating we get)

or,dy/dx×2y =-d/dx(x)^-2

or, dy/dx=2×(x)^-3÷2y=1/(x)^3×y

or, dy/dx=1/[(x^3)y] (please note that (1/(x^3y) is not correct)