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Implicit Differentiation

If x = t2...

Question

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

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Solution

$x = \frac{t^{2}}{1 + t^{2}}$

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

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

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

$∴ \frac{d x}{d t} = \frac{2 t}{{(1 + t^{2})}^{2}}$

$y = \frac{2 a t}{1 + t^{2}}$

$\frac{d y}{d t} = \frac{(1 + t^{2}) \frac{d}{d t} (2 a t) - (2 a t) \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}$

$\frac{d y}{d t} = \frac{(1 + t^{2}) (2 a) - (2 a t) (2 t)}{{(1 + t^{2})}^{2}}$

$∴ \frac{d y}{d t} = \frac{(1 + t^{2}) (2 a) - (2 a t) (2 t)}{{(1 + t^{2})}^{2}}$

$∴ \frac{d y}{d t} = \frac{(1 + t^{2} - 2 t^{2}) (2 a)}{{(1 + t^{2})}^{2}}$

$∴ \frac{d y}{d t} = \frac{(1 - t^{2}) (2 a)}{{(1 + t^{2})}^{2}}$

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

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x = \frac{1 - t^{2}}{1 + t^{2}}

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