assuming that ( x > 0 ), use differentiation to justify the formula ( int \frac { 1 } { x } d x = ln | x | +…

assuming that ( x > 0 ), use differentiation to justify the formula ( int \frac { 1 } { x } d x = ln | x | + c ).\nif ( int \frac { 1 } { x } d x = ln | x | + c ), then ( \frac { d } { d x } ( ln | x | + c ) = \frac { 1 } { x } ) for ( x > 0 ).\nthe derivative of ( ln | x | ) is ( square ) and the derivative of a constant ( c ) is ( square ), so ( int \frac { 1 } { x } d x = ln | x | + c ) for ( x > 0 ).

assuming that ( x > 0 ), use differentiation to justify the formula ( int \frac { 1 } { x } d x = ln | x | + c ).\nif ( int \frac { 1 } { x } d x = ln | x | + c ), then ( \frac { d } { d x } ( ln | x | + c ) = \frac { 1 } { x } ) for ( x > 0 ).\nthe derivative of ( ln | x | ) is ( square ) and the derivative of a constant ( c ) is ( square ), so ( int \frac { 1 } { x } d x = ln | x | + c ) for ( x > 0 ).

Answer

Explanation:

Step1: Derivative of ( \ln|x| )

For ( x>0 ), ( |x| = x ). The derivative of ( \ln(x) ) is ( \frac{1}{x} ) (by the formula ( \frac{d}{dx}(\ln u)=\frac{u'}{u} ), where ( u = x ) and ( u'=1 )).

Step2: Derivative of a constant

The derivative of a constant ( C ) is ( 0 ) (since ( \frac{d}{dx}(k)=0 ) for any constant ( k )).

Answer:

The derivative of ( \ln|x| ) is ( \frac{1}{x} ) and the derivative of a constant ( C ) is ( 0 ).