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Question Number 71729 by peter frank last updated on 19/Oct/19

find dU   if   U=x^2 e^(x/y)

finddUifU=x2exy

Commented by peter frank last updated on 19/Oct/19

thank you

thankyou

Commented by Prithwish sen last updated on 19/Oct/19

ylnu = 2yln(x) +x  du = [2xe^(x/y) dx+((x^2 e^(x/y) )/y) dx− (x^3 /y^2 )e^(x/y)  dy]   please check

ylnu=2yln(x)+xdu=[2xexydx+x2exyydxx3y2exydy]pleasecheck

Answered by mind is power last updated on 19/Oct/19

du=(∂u/∂x)dx+(∂u/∂y)dy  (∂/∂x)(x^2 e^(x/y) )=(2x+(x^2 /y))e^(x/y)   (∂/∂y)(x^2 e^(x/y) )=−(x^3 /y^2 )e^(x/y)   ⇒dU=(2x+(x^2 /y))e^(x/y) dx−(x^3 /y^2 )e^(x/y) dy

du=uxdx+uydyx(x2exy)=(2x+x2y)exyy(x2exy)=x3y2exydU=(2x+x2y)exydxx3y2exydy

Commented by peter frank last updated on 19/Oct/19

thank you

thankyou

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