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

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

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

Answer

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})