Finding the derivative of $y= (\ln x)^2$

derivatives

In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:

$y=(\ln x)^2$.
First, we take the power rule. This would make it $2(\ln x)$. Then you multiply it by the derivative of the inside function right? $\ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $\frac{2\ln x}x$. This doesn’t seem correct so I’m a little confused.

$y=\frac{3}{\sqrt{2x+1}}$

First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.

I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.

Best Answer

First one is correct, for the second one

\begin{eqnarray} \frac{{\rm d}}{{\rm d}x} \frac{3}{\sqrt{2x + 1}} &=& 3\frac{{\rm d}}{{\rm d}x} (2x + 1)^{-1/2} ~~~~\mbox{move constant out} \\ &=& -\frac{3}{2}(2x + 1)^{-3/2}\frac{{\rm d}}{{\rm d}x}(2x + 1) ~~~~\mbox{power rule + chain rule} \\ &=& -\frac{3}{2}(2x + 1)^{-3/2}(2)\\ &=& -\frac{3}{(2x + 1)^{3/2}} \end{eqnarray}