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प्रश्न
Find `bb(dy/dx)` for the given function:
yx = xy
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उत्तर
Given, yx = xy
Taking logarithm of both sides,
log yx = log xy
x log y = y log x ...(i)
Differentiating (i) w.r.t. x,
`x d/dx log y + log y d/dx (x)= y d/dx log x + log x d/dx y`
`=> x xx 1/y dy/dx + log y xx 1 = y xx 1/x + log x dy/dx`
`=> x/y dy/dx + log y = y/x + log x dy/dx`
`=> x/y dy/dx - log x dy/dx = y /x - log y`
`=> dy/dx (x/y - log x) = y /x - log y` ...(ii)
On multiplying both sides of (ii) by xy, we get
`=> dy/dx (x^2 - xy log x) = y ^2 - xy log y`
`therefore dy/dx = (y ^2 - xy log y)/(x^2 - xy log x)`
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