IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nknikhilesh1

X^y+Y^x=2 , find dy / dx?

mujtaba4iit

Re:please solve this question

allamraju

Diff on both sides,we get,

x^y[y/x+y'lnx]+y^x[lny+xy'/y]=0

y'=-[yx^y-1+y^xlny]/[x^ylnx+xy^x-1].

thunder2153

x^y+ y^x= 2

differntiate both sides w.r.t x

x^y ( dy/dx lnx + y/x ) + y^x (lny + dy/dx x/y)

x^y dy/dx lnx + x^{y - 1}y + y^x lny + y^x dy/dx x/y

dy/dx = -(x ^{y -1}y + y^xlny) / (x^y lnx + y^{x -1}y )

pundir

X^Y + Y^X = 2

YLOGX + XLOGY = LOG2

Y/X + LOGX*DY/DX + X/Y*DY/DX + LOGY = 0

DY/DX ( LOGX + X/Y ) = - ( Y/X + LOGY )

DY/DX { ( YLOGX + X ) /Y } = - { ( Y + XLOGY ) / X }

DY/DX = - Y (Y + XLOGY ) / X ( YLOGX + X )

Prajju

This can be more easily solved by partial differentiation method....
let f = x^y+y^x-2=0
differentiate f wrt x keeping y const.
df / dx = yx^(y-1) + y^xlogy - 0 = yx^(y-1) + y^xlog y ------- 1
differentiate f wrt y keeping x const....
df / dy = x^ylogx + xy^(x-1) ----------2
dividing 2 by 1
dy / dx = - { [ x^ylogx + xy^(x-1)] / [ yx^(y-1) + y^xlogy] }

Prajju

plz tell me wether my ans is correct or not.......

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Diff on both sides,we get,

xy[y/x+y'lnx]+yx[lny+xy'/y]=0

y'=-[yxy-1+yxlny]/[xylnx+xyx-1].

x^y[y/x+y'lnx]+y^x[lny+xy'/y]=0

y'=-[yx^y-1+y^xlny]/[x^ylnx+xy^x-1].