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I don’t get either of it since I didn’t get the chance to revise it, but the homework is due very soon, so I’ve been left with no choice :(

I don’t get either of it since I didn’t get the chance to revise it, but the homework-example-1
User CularBytes
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Given the function:


x^{(1)/(2)}\ln (3x)

We will differentiate with respect to x

For the given function, we will use the product rule


(f\cdot g)^(\prime)=f^(\prime)\cdot g+f\cdot g^(\prime)

we have two functions:


\begin{gathered} (x^{(1)/(2)})^(\prime)=\frac{1}{2x^{(1)/(2)}} \\ \ln (3x)^(\prime)=(1)/(3x)\cdot3=(1)/(x) \end{gathered}

So, the first derivative for the given function will be:


\begin{gathered} (x^{(1)/(2)}\ln \lbrack3x\rbrack)^(\prime)=\frac{1}{2x^{(1)/(2)}}\ln (3x)+x^{(1)/(2)}\cdot(1)/(x) \\ \\ =\frac{1}{2x^{(1)/(2)}}\ln (3x)+\frac{2x^{(1)/(2)}}{2x^{(1)/(2)}x^{(1)/(2)}} \\ \\ =\frac{1}{2x^{(1)/(2)}}\ln (3x)+\frac{2}{2x^{(1)/(2)}} \\ \\ =\frac{1}{2x^{(1)/(2)}}(\ln (3x)+2) \end{gathered}

User Shongsu
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