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Question

29. (x +secx) (x -tanx)

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Solution

Let, the function is,

f( x )=( x+secx )( xtanx )

Apply the formula of Leibnitz Product rule.

f ( x )=( x+secx ) d dx ( xtanx )+( xtanx ) d dx ( x+secx ) =( x+secx )[ d dx ( x ) d dx tanx ]+( xtanx )[ d dx ( x )+ d dx secx ] =( x+secx )[ 1 d dx tanx ]+( xtanx )[ 1+ d dx secx ]

Assume f 1 ( x )=tanx and f 2 ( x )=secx.

Replace x with ( x+h ), then f 1 ( x+h )=tan( x+h ) and f 2 ( x+h )=sec( x+h ).

Use first principal of derivative,

f 1 ( x )= lim h0 ( f 1 ( x+h ) f 1 ( x ) h ) = lim h0 ( tan( x+h )tan( x ) h ) = lim h0 ( tan( x+h )tanx h ) = lim h0 1 h ( sin( x+h ) cos( x+h ) sinx cosx )

Further simplify.

f 1 ( x )= lim h0 1 h ( sin( x+h )cosxsinxcos( x+h ) cos( x+h )cosx ) = lim h0 1 h ( sin( h ) cos( x+h )cosx ) =( lim h0 sinh h )( lim h0 1 cos( x+h )cosx )

Further simplify,

f 1 ( x )=1× 1 cos 2 x = sec 2 x

Again use first principal of derivative,

d dx tanx= sec 2 x(2)

f 2 ( x )= lim h0 ( f 2 ( x+h ) f 2 ( x ) h ) = lim h0 ( sec( x+h )sec( x ) h ) = lim h0 1 h ( 1 cos( x+h ) 1 cosx ) = lim h0 1 h ( cosxcos( x+h ) cos( x+h )cosx )

Further simplify,

f 2 ( x )= lim h0 1 h ( 2sin( x+x+h 2 )sin( xxh 2 ) cos( x+h ) ) = 1 cosx lim h0 1 h ( sin( 2x+h 2 )( sin( h 2 ) h 2 ) cos( x+h ) ) =secx ( lim h0 sin( 2x+h 2 ) )( lim h 2 0 sin( h 2 ) h 2 ) lim h0 cos( x+h )

Further simplify,

f 2 ( x )=secx sinx cosx =secxtanx d dx secx=sectanx (3)

Substitute the values from equation (2) and (3) into equation (1).

f ( x )=( x+secx )[ 1 sec 2 x ]+( xtanx )[ 1+secxtanx ]

Thus, derivative of ( x+secx )( xtanx ) is ( x+secx )[ 1 sec 2 x ]+( xtanx )[ 1+secxtanx ].


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