Given the function:
[tex]x^{\frac{1}{2}}\ln (3x)[/tex]
We will differentiate with respect to x
For the given function, we will use the product rule
[tex](f\cdot g)^{\prime}=f^{\prime}\cdot g+f\cdot g^{\prime}[/tex]
we have two functions:
[tex]\begin{gathered} (x^{\frac{1}{2}})^{\prime}=\frac{1}{2x^{\frac{1}{2}}} \\ \ln (3x)^{\prime}=\frac{1}{3x}\cdot3=\frac{1}{x} \end{gathered}[/tex]
So, the first derivative for the given function will be:
[tex]\begin{gathered} (x^{\frac{1}{2}}\ln \lbrack3x\rbrack)^{\prime}=\frac{1}{2x^{\frac{1}{2}}}\ln (3x)+x^{\frac{1}{2}}\cdot\frac{1}{x} \\ \\ =\frac{1}{2x^{\frac{1}{2}}}\ln (3x)+\frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}x^{\frac{1}{2}}} \\ \\ =\frac{1}{2x^{\frac{1}{2}}}\ln (3x)+\frac{2}{2x^{\frac{1}{2}}} \\ \\ =\frac{1}{2x^{\frac{1}{2}}}(\ln (3x)+2) \end{gathered}[/tex]
