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If y=x^x then prove that d2y/dx2 …

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If y=x^x then prove that d2y/dx2 - 1/y (dy/dx)^2 - y/x = 0

Posted by Pratik Roy 7 years, 10 months ago

CBSE > Class 12 > Mathematics

1 answers

Arka Ghosal 7 years, 10 months ago

Y=x^x Taking log on both sides log y=log x^x log y=x logx (1/y)(dy/dx)=(x*1/x)+logx (dy/dx)=y(1+log x) d/dx(dy/dx)=d/dx(y(1+logx)) d^2y/dx^2=(1+log x)(dy/dx)+y/x Now, 1/y(dy/dx)^2=(1/y)*y^2(1+logx)^2 Therefore d^2y/dx^2-1/y(dy/dx)^2-y/x=(1+log x)dy/dx - y(1+log x)^2 +y/x - y/x = (1+log x)(1+log x) * y - y(1+log x)^2 =y(1+log x)^2 - y(1+log x)^2 =0=RHS[PROVED]

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