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if x^{x}+x^{y}+y^{x}=a^{b}, then find `dy/dx`.

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#### Solution 1

x^{x}+x^{y}+y^{x}=a^{b........(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 = a^{b}

`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 = x^{y} and v = y^{x}

Then, u + v = ab

Differentiating both sides w.r.t x, we get

Concept: Logarithmic Differentiation

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