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Question
if xx+xy+yx=ab, then find `dy/dx`.
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Solution 1
xx+xy+yx=ab........(i)
`Let u=x^x`
`log u=xlogx`
`1/u*(du)/dx=x * 1/x+logx`
`therefore (du)/dx=x^x(1+logx)`
`Let v=x^y`
`logv =ylogx`
`1/v (dv)/dx=(y/x+logx dy/dx)`
`therefore (dv)/dx=x^y(y/x+logx dy/dx)`
`Let w=y^x`
`logw=x log y`
`1/w.(dw)/dx=(x/y*dy/dx+logy)`
`therefore (dw)/dx=y^x(logy+x/y*dy/dx)`
(i) can be written as
u + v + w = ab
`du/dx+dv/dx+dw/dx=0`
`=>x^x+x^xlogx+x^yy/x+x^y logx dy/dx+y^xlogy+y^x x/y dy/dx=0`
`=>dy/dx(x^ylogx+y^x x/y)=x^x+x^xlogx+x^y y/x+ y^x logy`
`=> dy/dx (x^y*logx+xy^(x-1))=(x^x+x^xlogx+yx^(y-1)+y^x*logy)`
`therefore dy/dx=(x^x+x^xlogx+yx^(y-1)+y^x*logy)/(x^y*logx+xy^(x-1))`
Solution 2
Let u = xy and v = yx
Then, u + v = ab
Differentiating both sides w.r.t x, we get
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