Question

If `x^7 * y^9 = (x + y)^16`, then show that `dy/dx = y/x`

Find `dy/dx` if x⁷ . y⁹ = (x + y)¹⁶.

Answer 1

x⁷ . y⁹ = (x + y)¹⁶

Taking logarithm of both sides, we get

log x⁷ . y⁹ = log (x + y)¹⁶

∴ log x⁷ + log y⁹ = 16 log (x + y)

∴ 7 log x + 9 log y = 16 log (x + y)

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

`7(1/x) + 9(1/y) dy/dx = 16(1/(x + y)) d/dx (x + y)`

∴ `7/x + 9/y dy/dx = 16/(x + y) (1 + dy/dx)`

∴ `7/x + 9/y dy/dx = 16/(x + y) + 16/(x + y) dy/dx`

∴ `9/y dy/dx - 16/(x + y) dy/dx = 16/(x + y) - 7/x`

∴ `(9/y - 16/(x + y)) dy/dx = 16/(x + y) - 7/x`

∴ `[(9x + 9y - 16y)/(y(x + y))] dy/dx = (16x - 7x - 7y)/(x(x + y))`

∴ `[(9x - 7y)/(y(x + y))] dy/dx = (9x - 7y)/(x(x + y))`

∴ `dy/dx = (9x - 7y)/(x(x + y)) xx (y(x + y))/(9x - 7y)`

∴ `dy/dx = y/x`