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If *y* = *x ^{x}*, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`

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#### Solution

y = x^{x}

Applying logarithm,

log y = x log x

`1/y dy/dx=logx+xxx1/x=1+logx`

`dy/dx=x^x[1+logx]`

`(d^2y)/dx^2=(d(x^2))/dx(1+logx)+x^x[d/dx(1+logx)]`

`=x^x(1+logx)(1+logx)+x^x[1/x]`

`=x^x(1+logx)^2+x^(x-1)`

`(d^2y)/dx^2-1/y(dy/dx)^2-y/x=x^x(1+logx)^2+x^(x-1)-1/(x^2)(x^x(1+logx)^2)-x^2/x`

= x^{x}(1+log x)^{2} + x^{x−1} −x^{x}(1+log x)^{2} − x^{x−1}

^{= 0}

Hence proved.

Concept: Simple Problems on Applications of Derivatives

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