Did I derive this correctly? More questions inside Sorry for the title, didn't quite know what to title it since I had a few questions. Anywho, I'm working on my homework (finding derivatives of the given function) and had a few questions.
$y=\ln\sqrt{5x+1}$ which of course equals $y=\ln\,(5x+1)^{1/2}$
$$g(x)=(5x+1)^{1/2} , g'(x)={1\over2}(5)^{-1/2}$$
$\displaystyle y'={g'(x)\over g(x)}$ which once I plug in turns to $\displaystyle{{1\over2}(5)^{-1/2}\over 2(5x+1)}$
Final answer: ${5\over2(5x+1)}$
I want to know why the ${1\over2}$ and $-{1\over2}$ exponent cancel out in the nominator and why the $2$ appears in the denominator. Does adding $(5x+1)^{1/2}$ in the denominator flip the exponent or am I in the wrong path.
I'm sorry for formatting, I tried using MathJax code but it doesn't look like im using it correctly.
Also, could someone explain how $f(x)=e^{x^3/3}$ equals $f'(x)=x^2 e^{x^3/3}$? I'm following the chain rule but I guess I just want to understand it better?
 A: These are all examples of the chainrule $$h'(x)=f'(g(x))g'(x)$$ where $h(x)=f(g(x))$. Simply take the derivative (not "derive", that is used for writing proofs) of the outer function $f(x)$, plug the inner function $g(x)$ into that and multiply by the derivative of the inner function. Let's take an example:
$$y=\frac{1}{2}\ln(5x+1),  f(x)=\frac{1}{2}\ln(x), g(x)=5x+1 \implies y'=\frac{1}{2}\frac{1}{5x+1}5$$
Do you see why? 
Similarly with $g(x)$, where the outer function is $\sqrt{x}$ and the inner is $5x+1$. You can't just use the usual rule for polynomial terms $\left(ax^n \right)'=nax^{n-1}$ since you instead of just $x$ (as it is written in the rule) now have a polynomial, $5x+1$, and then you need to use the chain rule. 
Try to use this information to calculate the rest of the derivatives and then let me know if you're still having trouble with it. 
A: As Friedrich Philipp said, you must use the chain rule correctly, because your derivative is not correct. If your $g(x)$ is $(5x+1)^{1/2}$, then you have to apply the chain rule once more to find the derivative. By the chain rule, 
$$g'(x) = \frac{1}{2}(5x+1)^{-1/2}\cdot (5x + 1)'=\frac{5}{2}(5x + 1)^{-1/2}$$
As a side note, you can tell that the derivative is wrong because when you found $g'(x)$, you found $\frac{1}{2}\cdot 5^{-1/2}$, which is a constant. Only linear functions will differentiate into constants, so that definitely cannot be correct. 
The reason why $f'(x) = x^2 e^{x^3/3}$ is because of the chain rule. By the chain rule, $(f(g(x)))'=f'(g(x))g'(x)$. When you let $f(x)=e^{x}$ and $g(x)=x^3/3$, you get the desired answer:
$$\left(e^{x^3/3}\right)'=f'(x^3/3)(x^3/3)'=e^{x^3/3}\cdot x^2$$
A: As mentioned in the comments, part of computing the derivative in your post looks incorrect. Here is what I see as an answer:
Given:
$y=\ln\sqrt{5x+1}$
You can write this as:
$$y(x)=ln (g(x))$$
Where $$g(x)=\sqrt{5x+1}$$
Using chain rule, 
$$ dg/dx=(1/2)\frac{d/dx(5x+1)}{\sqrt{5x+1}}$$
$$ g'(x)=(1/2)\frac{5}{\sqrt{5x+1}}$$
Since $y' (x) = \frac{g'(x)}{g(x)}$
$$y'(x) = (1/2) \frac{\frac{5}{\sqrt{5x+1}}}{\sqrt{5x+1}}$$
Simplifying:
$$y'(x) = \frac{5}{2(5x+1)}$$
