Question

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

Answer 1

x² + 6xy + y² = 10        ...(i)

Differentiating both sides w.r.t. x, we get

`2x + 6(x ("d"y)/("d"x) + y) + 2y ("d"y)/("d"x)` = 0

∴ `2x + 6x  ("d"y)/("d"x) + 6y + 2y ("d"y)/("d"x) = 0`

∴ `(2x + 6y) + (6x + 2y) ("d"y)/("d"x) = 0`

∴ `("d"y)/("d"x) = - (x + 3y)/(3x + y)`           ...(ii)

∴ (3x + y) `("d"y)/("d"x)` = − (x + 3y)

Again, differentiating both sides w.r.t. x, we get

`(3x + y) ("d"^2y)/("d"x^2) + ("d"y)/("d"x) (3 + ("d"y)/("d"x)) = - (1 + 3 * ("d"y)/("d"x))`

∴ `3 ("d"y)/("d"x) + (("d"y)/("d"x))^2 + 1 + 3("d"y)/("d"x) = - ("d"^2y)/("d"x^2)`(y + 3x)

∴ `(("d"y)/("d"x))^2 + 6 ("d"y)/("d"x) + 1 = - ("d"^2y)/("d"x^2)`(y + 3x)

∴ `[- ((x + 3y)/(3x + y))]^2 + 6 [(- (x + 3y))/(3x + y)] + 1`

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

By solving, we get

`(x^2 + 9y^2 + 6xy - 6xy - 18x^2 - 18y^2 - 54xy + y^2 + 9x^2 + 6xy)/(y + 3x)^2 = - ("d"^2y)/("d"x^2)`(y + 3x)

∴ `- ("d"^2y)/("d"x^2)  (y + 3x)^3 = - 8x^2 - 8y^2 - 48xy`

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

= −8 × 10          ...[from (i)] 

= −80

∴ `("d"^2y)/("d"x^2) = 80/(3x + y)^3`