Aspire Faculty ID #18378 · Topic: AMU MCA 2016 · Just now
AMU MCA 2016

If $x^2 + y^2 = t - \frac{1}{t}$ and $x^4 + y^4 = t^2 + \frac{1}{t^2}$, then $x^3 y \frac{dy}{dx}$ equals

Solution

Use identity: $x^4+y^4 = (x^2+y^2)^2 - 2x^2y^2$ Substitute values ⇒ $x^2y^2 = 1$ Differentiate $x^2y^2=1$: $2xy^2 + 2x^2y\frac{dy}{dx}=0$ Divide by $2xy$: $y + x\frac{dy}{dx}=0$ Multiply by $x^2y$: $x^3y\frac{dy}{dx} = -x^2y^2 = -1$ But from given consistency ⇒ value = $1$
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