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Question
Find `bb(dy/dx)` for the given function:
xy + yx = 1
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Solution
xy + yx = 1 ....(i)
Differentiating (i) w.r.t. x we get
`d/dx (x^y) + d/dx (y^x)` = 0 ...(ii)
Let u = xy
Taking log on both sides, we get
log u = y log x ....(iii)
Differentiating the above w.r.t. x, we get
`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]` ...(iv)
Let v = yx
⇒ log v = x log y ... (v)
Differentiating the above w.r.t. x, we get
`1/v (dv)/dx = d/dx x log y`
= `log y d/dx (x) + x d/dx (log y)`
= `log y xx 1 + x xx 1/y dy/dx`
= `log y + x/y dy/dx`
`therefore (dv)/dx = v[log y + x/y dy/dx]`
= `y^x [log y + x/y dy/dx]` .....(vi)
Substituting the values of (iv) and (vi) in (ii), we get
`x^y [y/x + log x dy/dx] + y^x [log y + x/y dy/dx] = 0`
`(x^y log x + xy^(x - 1)) dy/dx = - (y^x log y + yx^(y - 1))`
`dy/dx = -(y^x log y + yx^(y - 1))/(x^y log x + xy^(x - 1))`
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