Question

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

Answer 1

We have,

$x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$

$\Rightarrow (x^2+y^2{)}^2=(t-\frac{1}{t}{)}^2$

$\Rightarrow x^4+y^4+2x^2{y}^2=t^2+\frac{1}{t^2}-2$

$\Rightarrow x^4+y^4+2x^2{y}^2=x^4+y^4-2$ [given]

$\Rightarrow 2x^2{y}^2=-2$

$\Rightarrow x^2{y}^2=-1$

$\Rightarrow y^2=-\frac{1}{x^2}$

$\Rightarrow y^2=-x^{-2}$

Differentiating w.r.t. x, we get,

$\Rightarrow 2y\frac{dy}{dx}=-(-2)x^{-3}$

$\Rightarrow y\frac{dy}{dx}=\frac{1}{x^3}$

$\therefore \frac{dy}{dx}=\frac{1}{x^{3}y}$