First slide

Derivatives

Question

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

Moderate

A

$- \frac{x}{y}$
B

$- \frac{y}{x}$
C

$\frac{x^{2}}{y^{2}}$
D

$\frac{y^{2}}{x^{2}}$

Solution

$x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}} = {(t + \frac{1}{t})}^{2} - 2 = {(x^{2} + y^{2})}^{2} - 2 = x^{4} + y^{4} + 2 x^{2} y^{2} - 2$
$\Rightarrow 2 x^{2} y^{2} = 2 \Rightarrow x^{2} y^{2} = 1 \Rightarrow 2 x^{2} y \frac{d y}{d x} + 2 x y^{2} = 0 \Rightarrow x \frac{d y}{d x} + y = 0 \Rightarrow \frac{d y}{d x} = \frac{- y}{x} .$

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