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Question
If `x^3y^5 = (x + y)^8` , then show that `(dy)/(dx) = y/x`
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Solution
Given `x^3y^5 = (x + y)^8`
Taking log on both sides
3 logx + 5 logy = 8 log(x + y)
Differentiate w.r.t. x
`3/x + 5/y (dy)/(dx) = 8(1/(x + y))[1 + (dy)/(dx)]`
`5/y (dy)/(dx) - 8/(x + y) dy/dx = 8/(x + y) - 3/x`
`(dy)/(dx)[5/y - 8/(x + y)] = (8x - 3x - 3y)/(x(x + y))`
`(dy)/(dx)[[5x + 5y - 8y]/(y(x + y))] = (5x - 3y)/(x(x + y))`
`(dy)/(dx) = [(5x - 3y)/(x(x + y)]] xx [(y(x + y))/(5x - 3y)]`
`(dy)/(dx) = y/x`
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