# If x2 + 6xy + y2 = 10, then show that d2ydx2=80(3x+y)3 - Mathematics and Statistics

Sum

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

#### Solution

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

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Chapter 3: Differentiation - Miscellaneous Exercise 3 [Page 101]

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Differentiation
Miscellaneous Exercise 3 | Q 4.22 | Page 101

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