Question

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

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 \( 0 \).