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Question
Find `dy/dx`for the function given in the question:
xy + yx = 1
Solution
Given, ∵ xy + yx = 1
Let, u = xy, v = yx ∵ u + v = 1
Differentiating both sides with respect to x,
`1/u (du)/dx = d/dx y log x`
`= y d/dx log x +log x d/dx (y)`
`= y * 1/x + log x * dy/dx = y/x + log x dy/dx`
`therefore (du)/dx = u [y/x + log x dy/dx]`
`= x^y [y/x + log x dy/dx]` ...(2)
Now, v = yx
Taking logarithm of both sides, log, log v = log yx = x log y
Differentiating both sides with respect to ,
`1/v (dv)/dx = d/dx x log y`
`= x d/dx log y +log y d/dx (x)`
`= x * 1/y + log yxx 1 = x/y dy/dx + log y`
`therefore (dv)/dx = v [x/y dy/dx + log y]`
`= y^x [x/y dy/dx + log y]` ...(3)
From equation (2) and (3), putting the values of `(du)/dx` and `(dv)/dx` in equation (1),
`therefore x^y [(log x) dy/dx + y/x] + y^x [log y + x/y dy/dx] = 0`
`therefore (x^y log x + xy^x/y) dy/dx + x^y * y/x + y^x log y = 0`
`therefore (x^y log x + x y^(x - 1)) dy/dx + yx^(y - 1) + y^x log y = 0`
`therefore dy/dx = (yx^(y - 1) + y^x log y)/(x^y log x + x y^(x - 1))`
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