If y = x^x, prove that (d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0. - Mathematics

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

 
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Solution

y = xx

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`

= xx(1+log x)2 + xx1 xx(1+log x)2  xx1

= 0

Hence proved.

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2013-2014 (March) Delhi Set 1

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