Question

11·(x cos x)" + (x sin x)"

Answer 1

The given function is ( xcosx ) x + ( xsinx ) 1 x .

Let y= ( xcosx ) x + ( xsinx ) 1 x .

Consider,

u= ( xcosx ) x (1)

And,

v= ( xsinx ) 1 x (2)

Now, y=u+v.

Differentiate both sides with respect to x.

dy dx = d( u+v ) dx dy dx = du dx + dv dx (3)

Take log on both sides of the equation (1).

logu=log ( xcosx ) x logu=xlog( xcosx ) logu=x( logx+logcosx ) logu=xlogx+xlogcosx

Differentiate both sides with respect to x.

d( logu ) dx = d( x.logx+x.logcosx ) dx 1 u du dx = d( x.logx ) dx + d( x.logcosx ) dx 1 u . du dx =( d( x ) dx .logx+ d( logx ) dx .x )+( d( x ) dx .logcosx+ d( logcosx ) dx .x ) 1 u . du dx =( 1.logx+( 1 x ).x )+( 1.logcosx+( −sinx cosx ).x )

Further simplify.

1 u . du dx =( logx+1 )+( logcosx+( −sinx cosx ).x ) 1 u . du dx =( logx+1 )+( logcosx−xtanx ) du dx =u[ ( logx+1 )+( logcosx−xtanx ) ]

Substitute u= ( xcosx ) x in the above equation.

du dx = ( xcosx ) x [ ( logx+1 )+( logcosx−xtanx ) ] du dx = ( xcosx ) x [ 1−xtanx+logx+logcosx ] du dx = ( xcosx ) x [ 1−xtanx+log( xcosx ) ] (4)

Take log on both sides of equation (2).

logv=log( ( xsinx ) 1 x ) logv= 1 x .log( xsinx ) logv= 1 x .( logx+logsinx ) logv= 1 x logx+ 1 x logsinx

Differentiate both sides with respect to x.

d( logv ) dx = d( 1 x logx+ 1 x logsinx ) dx 1 v ( dv dx )= d( 1 x logx ) dx + d( 1 x logsinx ) dx 1 v . dv dx =( 1 x d( logx ) dx +logx d( 1 x ) dx )+( 1 x d( logsinx ) dx +logsinx d( 1 x ) dx ) 1 v . dv dx =( 1 x ⋅ 1 x +logx( − 1 x 2 ) )+( 1 x ⋅ cosx sinx +logsinx( − 1 x 2 ) )

Further simplify.

1 v . dv dx =( 1−logx x 2 )+( 1 sinx cosx.x−logsinx x 2 ) 1 v . dv dx =( 1−logx x 2 )+( cotx.x−logsinx x 2 ) dv dx =v[ ( 1−logx x 2 )+( cotx.x−logsinx x 2 ) ]

Substitute v= ( xsinx ) 1 x in the above equation.

dv dx = ( xsinx ) 1 x ( 1−logx+xcotx−logsinx x 2 ) dv dx = ( xsinx ) 1 x ( 1+xcotx−( logx+logsinx ) x 2 ) dv dx = ( xsinx ) 1 x ( 1+xcotx−log( xsinx ) x 2 ) (5)

By substituting the value of du dx from equation (4) and dv dx from equation (5) in equation (3), we get

dy dx = ( xcosx ) x [ 1−xtanx+log( xcosx ) ]+ ( xsinx ) 1 x ( 1+xcotx−log( xsinx ) x 2 )