Question

Differentiate the function with respect to x .

Answer 1

Given expression is ( x+ 1 x ) x + x ( 1+ 1 x ) .

Let the expression be y= ( x+ 1 x ) x + x ( 1+ 1 x ) .

Let, u= ( x+ 1 x ) x and v= x ( 1+ 1 x ) .

So, y=u+v.

Differentiate both sides of the expression with respect to x.

dy dx = d( u+v ) dx dy dx = du dx + dv dx (1)

Calculate du dx , where u= ( x+ 1 x ) x .

Take log on both sides of u.

logu=log ( x+ 1 x ) x

As we know that log( a b )=bloga,

logu=xlog ( x+ 1 x )

Now, differentiate the expression on both sides with respect to x.

d( logu ) dx ( du du )= d( xlog( x+ 1 x ) ) dx 1 u ( du dx )= d( x ) dx .log( x+ 1 x )+ d( log( x+ 1 x ) ) dx .x 1 u ( du dx )=1.log( x+ 1 x )+( ( 1 x+ 1 x ). d dx ( x+ 1 x ) ).x 1 u ( du dx )=log( x+ 1 x )+( 1 x+ 1 x .( d( x ) dx + d( 1 x ) dx ) ).x (2)

Further simplify the above equation (2).

1 u ( du dx )=log( x+ 1 x )+( 1 x+ 1 x .( 1+ −1 x 2 ) ).x 1 u ( du dx )=log( x+ 1 x )+( x x 2 +1 .( x 2 −1 x 2 ) ).x 1 u ( du dx )=log( x+ 1 x )+( x 2 −1 x 2 +1 ) ( du dx )=u( log( x+ 1 x )+( x 2 −1 x 2 +1 ) ) (3)

Substitute the value of u= ( x+ 1 x ) x in equation (3).

( du dx )= ( x+ 1 x ) x ( log( x+ 1 x )+( x 2 −1 x 2 +1 ) )

Now, calculate dv dx , where v= x ( 1+ 1 x ) .

Take log on both sides of u.

logv=log ( x+ 1 x ) x

As we know that log( a b )=bloga,

logv=log x ( 1+ 1 x ) logv=( 1+ 1 x )logx

Now, differentiate the expression on both sides with respect to x.

d( logv ) dx ( dv dv )= d( ( 1+ 1 x ).logx ) dx d( logv ) dv ( dv dx )= d( ( 1+ 1 x ).logx ) dx 1 v ( dv dx )= d( ( 1+ 1 x ).logx ) dx (4)

Apply product rule in ( x+ 1 x ).logx and simplify the equation (4).

1 v ( dv dx )= d( 1+ 1 x ) dx .logx+ d( logx ) dx .( 1+ 1 x ) 1 v ( dv dx )=( d( 1 ) dx + d( 1 x ) dx ).logx+ d( logx ) dx .( 1+ 1 x ) 1 v ( dv dx )=( 0+( −1 x 2 ) ).logx+ 1 x .( 1+ 1 x ) 1 v ( dv dx )=( −1 x 2 ).logx+ 1 x .( 1+ 1 x ) (5)

Further simplify the equation (5).

1 v ( dv dx )= −logx x 2 + 1 x + −1 x 2 1 v ( dv dx )=( −logx+x+1 x 2 ) dv dx =v( −logx+x+1 x 2 ) (6)

Substitute the value of v= x ( 1+ 1 x ) in equation (6).

dv dx = x ( 1+ 1 x ) ( x+1−logx x 2 )

Now, substitute the value of du dx & dv dx in equation (1).

dy dx =( du dx )= ( x+ 1 x ) x ( log( x+ 1 x )+( x 2 −1 x 2 +1 ) )+ x ( 1+ 1 x ) ( x+1−logx x 2 )

Thus, the solution is,

dy dx =( du dx )= ( x+ 1 x ) x ( log( x+ 1 x )+( x 2 −1 x 2 +1 ) )+ x ( 1+ 1 x ) ( x+1−logx x 2 )