Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

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

Solution:

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

Differentiating it with respect to $t$ using quotient rule,

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

$=\left[\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$=\left[\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}}\right]$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{-4 \mathrm{t}}{\left(1+\mathrm{t}^{2}\right)^{2}}\right] \ldots . . .(1)$

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

Differentiating it with respect to $t$ using quotient rule,

$\frac{d y}{d t}=\left[\frac{\left(1+t^{2}\right) \frac{d}{d t}(2 t)-(2 t) \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right]$

$=\left[\frac{\left(1+t^{2}\right)(2)-(2 t)(2 t)}{\left(1+t^{2}\right)^{2}}\right]$

$=\left[\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}\right]$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2\left(1-\mathrm{t}^{2}\right)}{\left(1+\mathrm{t}^{2}\right)^{2}} \ldots \ldots(2)$

divided equation (2)by (1) so,

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \times \frac{1}{\frac{-4 t}{\left(1+t^{2}\right)^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\left(1-\mathrm{t}^{2}\right)}{-4 \mathrm{t}}$

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