If x2-y2-t-1-t and x4-y4-t2-1-t2- then prove that d y-d x-y-x

We have, nbsp;x^2+y^2=t+1t Squaring both sides, we get ⇒ nbsp;(x^2+y^2)^2=(t+1t)^2 x^4+y^4+2x^2y^2=t^2+1t^2+2 x^4+y^4+2x^2y^2=x^4+y^4+2 2x^2y^2=2 ⇒ x^2y^2=1 ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

If x2+y2=t+1/...

Question

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

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Solution

We have, $x^{2} + y^{2} = t + \frac{1}{t}$
Squaring both sides, we get
$\Rightarrow (x^{2} + y^{2})^{2} = {(t + \frac{1}{t})}^{2}$
$x^{4} + y^{4} + 2 x^{2} y^{2} = t^{2} + \frac{1}{t^{2}} + 2$
$x^{4} + y^{4} + 2 x^{2} y^{2} = x^{4} + y^{4} + 2$
$2 x^{2} y^{2} = 2 \Rightarrow x^{2} y^{2} = 1$
Differentiating with respect to $x$ ,
$x^{2} \times 2 y \frac{d y}{d x} + y^{2} \times 2 x = 0$
$\frac{d y}{d x} = \frac{- y^{2} \times 2 x}{x^{2} \times 2 y} = - \frac{y}{x}$ (proved)

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If x2+y2=t+1t and x4+y4=t2+1t2, then prove that dydx=−yx

We have, x2+y2=t+1t Squaring both sides, we get ⇒(x2+y2)2=(t+1t)2 x4+y4+2x2y2=t2+1t2+2 x4+y4+2x2y2=x4+y4+2 2x2y2=2⇒x2y2=1 Differentiating with respect to x, x2×2ydydx+y2×2x=0 dydx=−y2×2xx2×2y=−yx (proved)

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