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प्रश्न
`sin xy + x/y` = x2 – y
उत्तर
Given that: `sin xy + x/y` = x2 – y
Differentiating both sides w.r.t. x
`"d"/"dx" sin(xy) + "d"/"dx"(x/y) = "d"/"dx" (x^2) - "d"/"dx"(y)`
⇒ `cos xy * "d"/"dx" (xy) + (y * "d"/"dx" * x - x * "dy"/"dx")/y^2 = 2x - "dy"/"dx"`
⇒ `cos y [x * "dy"/"dx" + y * 1] + ("y"*1)/"y"^2 - x/y^2 * "dy"/"dx" = 2x - "dy"/"dx"`
⇒ `x cos xy * "dy"/"dx" + y cos xy + 1/y - x/y^2 "dy"/"dx" = 2x - "dy"/"dx"`
⇒ `x cos xy * "dy"/"dx" - x/y^2 * "dy"/"dx" + "dy"/"dx" = -y cos xy - 1/y + 2x`
⇒ `[x cos xy - x/y^2 + 1] "dy"/"dx" = 2x - y cos xy - 1/y`
⇒ `([xy^2 cos xy - x + y^2])/y^2 "dy"/"dx" = (2xy - y^2 cos xy - 1)/y`
⇒ `"dy"/"dx" = (2xy - y^2 cos xy - 1)/y xx y^2/(xy^2 cos xy - x + y^2)`
= `(2xy^2 - y^3 cos(xy) - y)/(xy^2 cos (xy) - x + y^2)`
Hence, `"dy"/"dx" = (2xy^2 - y^3 cos(xy) - y)/(xy^2 cos (xy) - x + y^2)`.
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