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# Find of function.

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## The given function is x y + y x =1. Let u= y x (1) And v= x y (2) So, u+v=1. Differentiate both sides with respect to x. d( u+v ) dx = d( 1 ) dx du dx + dv dx =0 (3) Take log on both sides of equation (1). logu=log ( y ) x logu=xlogy Differentiate both sides of the above equation with respect to x. d( logu ) dx = d( x.logy ) dx 1 u du dx = d( x ) dx .logy+ d( logy ) dx .x 1 u du dx =logy+x. 1 y . dy dx du dx =u( logy+ x y . dy dx ) Substitute u= y x in the above equation. du dx = y x ( logy+ x y . dy dx ) du dx = y x logy+ y x−1 .x dy dx (4) Take log on both sides of equation (2). logv=log( x y ) logv=y.logx Differentiate both sides with respect to x. d( logv ) dx = d( ylogx ) dx 1 v ( dv dx )= d( y ) dx .logx+ d( logx ) dx .y 1 v ( dv dx )= dy dx .logx+ 1 x .y dv dx =v( dy dx .logx+ y x ) Substitute v= x y in the above equation. dv dx = x y ( dy dx .logx+ y x ) dv dx = x y .logx dy dx + x y . y x (5) By substituting the value of du dx from equation (4) and dv dx from equation (5) in equation (3), we get ( y x logy+ y x−1 .x dy dx )+( x y .logx dy dx + x y . y x )=0 dy dx ( x y logy+x y x−1 )=−( x y−1 ×y+ y x .logy ) dy dx =− ( y x y−1 + y x logy ) ( x y logy+x y x−1 ) Thus, differentiation of x y + y x =1 is dy dx =− ( y x y−1 + y x logy ) ( x y logy+x y x−1 ) .

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