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Question

Find of function.

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Solution

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 x1 .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 x1 .x dy dx )+( x y .logx dy dx + x y . y x )=0 dy dx ( x y logy+x y x1 )=( x y1 ×y+ y x .logy ) dy dx = ( y x y1 + y x logy ) ( x y logy+x y x1 )

Thus, differentiation of x y + y x =1 is dy dx = ( y x y1 + y x logy ) ( x y logy+x y x1 ) .


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