How do you implicitly differentiate #-y=xy-e^ysqrt(x-2) #?

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Answer 1 · 2016-09-05T12:56:05

#dy/dx=(y(x-5))/{2(x+1)(x-2)(y-1)}#.

Let us rewrite the given eqn. as, # : e^ysqrt(x-2)=xy+y=y(x+1)#.

#:. ln(e^ysqrt(x-2))=ln{y(x+1)}#.

#:. lne^y+ln(x-2)^(1/2)=lny+ln(x+1)#

#:. y+1/2ln(x-2)=lny+ln(x+1)#.

#:. d/dx{y+1/2ln(x-2)}=d/dx{lny+ln(x+1)}#.

#:. dy/dx+1/2d/dxln(x-2)=d/dx(lny)+d/dxln(x+1)#.

#:. dy/dx+1/2*1/(x-2)=d/dylny*dy/dx+1/(x+1)......"[Chain Rule]"#.

#:. dy/dx+1/(2(x-2))=1/y*dy/dx+1/(x+1#.

#:. (1-1/y)dy/dx=1/(x+1)-1/(2(x-2))#.

#:. (y-1)/y*dy/dx={2(x-2)-(x+1)}/{2(x+1)(x-2)}#.

#:. dy/dx=(y(x-5))/{2(x+1)(x-2)(y-1)}#.

Enjoy Maths.!