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प्रश्न
If `x^7 * y^9 = (x + y)^16`, then show that `dy/dx = y/x`
Find `dy/dx` if x7 . y9 = (x + y)16.
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उत्तर
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`
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