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Question
Let z(x, y) = x tan–1(xy), x = t², y = s et, s, t ∈ R. Find `(delz)/(del"s")` and `(delz)/(del"t")` at s = t = 1
Solution
`(delz)/(del"z") = x 1/(1 + (xy)^2) xx y + tan^-1 (xy)`
`(xy)/(1 + x^2y^2) + tan^-1 (xy)`
`(delz)/(dely) = x 1/(1 + (xy)^2) xx x`
= `x^2/(1 + x^2y^2)`
`(delx)/(dely) = 0, (dely)/(del"s") = "e"^"t", (delx)/(del"t") = 2"t", (dely)/(del"t") = "s" "e"^"t"`
`(delz)/(del"s") = (delz)/(del"s") + (delz)/(dely) (dely)/(del"s")`
`(delz)/(del"s") = ([(xy)/( + x^2y^2)] + tan^-1 (xy)) xx 0 + x^2/(1 + x^2y^2) xx "e"^"t"`
= `"t"^4/(1 + "t"^2"s"^2"e"^(2"t")) "e"^"t" = ("e"^"t" "t"^4)/(1 + "t"^2"s"^2"e"^(2"t"))`
`(delz)/(del"t") = (delz)/(del"x") (delx)/(delt) + (delz)/(del"y") (dely)/(del"t")`
= `[(xy)/(1 + (xy)^2) + tan^-1(xy)] xx 2"t" + [x^2/(1 + (xy)^2)] xx "se"^"t"`
= `[("t"^2"se"^"t")/(1 + "t"^2"s"^2"e"^(2"t")) + tan^-1("t"^2"se"^"t")] xx 2"t" + ["t"^4/(1 + "t"^2"s"^2"e"^(2"t"))]"se"^"t"`
As s = t = 1,
`(delz)/(del"s") = "e"/(1 + "e"^2)`
`(delz)/(del"t") = (3"e")/(1 + "e"^2) + 2tan^-1("e")`
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