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Let z(x, y) = x tan–1(xy), x = t², y = s et, s, t ∈ R. Find s∂z∂s and t∂z∂t at s = t = 1 - Mathematics

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प्रश्न

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

योग
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उत्तर

`(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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Linear Approximation and Differential of a Function of Several Variables
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differentials and Partial Derivatives - Exercise 8.6 [पृष्ठ ८४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 8 Differentials and Partial Derivatives
Exercise 8.6 | Q 6 | पृष्ठ ८४

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