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प्रश्न
If x = `(t + 1)/(t - 1), y = (t - 1)/(t + 1), "then show that" y^2 + "dy"/"dx"` = 0.
उत्तर
x = `("t" + 1)/("t" - 1)` .......(1)
y = `("t" - 1)/("t" + 1)` ......(2)
Multiplying equations (1) and (2)
x × y = `(("t" + 1))/(("t" - 1)) xx (("t" - 1))/(("t" + 1))`
x × y = 1
x = `1/"y"` ....(3)
Differentiating w.r.t. x,
`"x" xx "dy"/"dx" + "y" xx (1) = 0`
`1/"y" xx "dy"/"dx" + "y"/1 = 0` ....[From (3)]
`(1 xx "dy"/"dx" + "y"^2)/"y" = 0`
`"dy"/"dx" + "y"^2 = 0`
∴ `"y"^2 + "dy"/"dx" = 0`
Hence proved.
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