Question

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

Explanation:

Step1: Differentiate \( \ln|x| \)

Since \( x>0 \), \( |x| = x \). The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{d}{dx}(\ln(x))=\frac{1}{x} \).

Step2: Differentiate the constant \( C \)

The derivative of a constant \( C \) with respect to \( x \) is \( \frac{d}{dx}(C) = 0 \).

Step3: Use the sum rule of differentiation

By the sum rule \( \frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx} \), where \( u=\ln|x| \) and \( v = C \). So \( \frac{d}{dx}(\ln|x|+C)=\frac{d}{dx}(\ln|x|)+\frac{d}{dx}(C) \).

Substituting the results from Step1 and Step2, we get \( \frac{d}{dx}(\ln|x|+C)=\frac{1}{x}+0 \).

Answer:

\(\frac{1}{x}\)