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X = ttt+1t, y = ttt-1t

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प्रश्न

x = `"t" + 1/"t"`, y = `"t" - 1/"t"`

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उत्तर

Given that,

x = `"t" + 1/"t"`, y = `"t" - 1/"t"`

Differentiating both the given parametric functions w.r.t. t

`"dx"/"dt" = 1 - 1/"t"^2`, `"dy"/"dt" = 1 + 1/"t"^2`

∴ `"dy"/"dx" = (("dy")/("dt"))/(("dx")/("dt"))`

= `(1 + 1/"t"^2)/(1 - 1/"t"^2)`

= `("t"^2 + 1)/("t"^2 - 1)`

Hence, `"dy"/"dx" = ("t"^2 + 1)/("t"^2 - 1)`.

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अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ ११०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 44 | पृष्ठ ११०

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