Question

If x² + 6xy + y² = 10, show that `(d^2y)/(dx^2) = (80)/(3x + y)^3`.

Answer 1

x² + 6xy + y² = 10                              ...(1)
Differentiating both sides w.r.t. x, we get

`2x + 6[x"dy"/"dx" + y."d"/"dx"(x)] + 2y"dy"/"dx"` = 0

∴ `2x + 6x"dy"/"dx" + 6y xx 1 + 2y"dy"/"dx"` = 0

∴ `(6x + 2y)"dy"/"dx"` = – 2x – 6y

∴ `"dy"/"dx" = (-2(x + 3y))/(2(3x + y)) = -((x + 3y)/(3x + y))`                ...(2)

∴ `(d^2y)/(dx^2) = -"d"/"dx"((x + 3y)/(3x + y))`

= `-[((3x + y)."d"/"dx"(x + 3y) - (x + 3y)."d"/"dx"(3x + y))/(3x + y)^2]`

= `-[((3x + y)(1 + 3"dy"/"dx") - (x + 3y)(3 + "dy"/"dx"))/(3x + y)^2]`

= `(1)/(3x + y)^2[-(3x + y){1 - (3(x + 3y))/(3x + y)} + (x + 3y)(3 - (x + 3y)/(3x + y))]`         ...[By (2)]

= `(1)/(3x + y)^2[-(3x + y)[-(3x + y)((3x + y - 3x - 9y)/(3x + y)) + (x + 3y)((9x + 3y - x - 3y)/(3x + y))]`

= `(1)/(3x + y)^2[8y + ((x + 3y)(8x))/(3x + y)]`

= `(1)/(3x + y)^2[((8y(3x + y) + (x + 3y)8x))/(3x + y)]`

= `(24xy + 8y^2 + 8x^2 + 24xy)/(3x + y)^2`

= `(8x^2 + 48xy + 8y^2)/(3x + y)^3`

= `(8(x^2 + 6xy + y^2))/(3x + y)^3`

= `(8(10))/(3x + y)^3`                         ...[By (1)]

∴ `(d^2y)/(dx^2) = (80)/(3x + y)^3`.