inverse derivative shortcut and chain rule I know that when you have the equation y=ln(x), and you need to find the derivative, you can use the shortcut y'= 1/x.  My question is why, when using the shortcut, do you have to multiply by the derivative of x?  I'm aware it has something to do with the chain rule but I don't understand why.
Here's an example of my question:  If y=ln(5x), why doesn't y'=1/5x ?    
 A: If $y = \ln x$ then $\dfrac{dy}{dx} = \dfrac 1 x$.  That's not a "shortcut"; that is the derivative of this function.
Recall that $\ln(5x) = \ln 5 + \ln x$, so $\dfrac d {dx}\ln(5x) = \dfrac d{dx} \ln 5 + \dfrac d {dx} \ln x$, and $\dfrac d {dx}\ln 5 = 0$ since $\ln 5$ is a constant, i.e. $\ln 5$ does not change as $x$ changes.
The chain rule can be stated in the form $\dfrac{dy}{dx} = \dfrac{dy}{du}\,\dfrac{du}{dx}$.  For example, if we have $y=(x^3+8x+5)^{36}$, we can write
\begin{align}
y = u^{36}, & & u = x^3 + 8x+5 \\[10pt]
\frac{dy}{du} = 36u^{35}, & & \frac{du}{dx} = 3x^2 + 8,
\end{align}
so
$$
\frac{dy}{dx} = 36u^{35}\cdot (3x^2+8) = 36(x^3+8x+5)^{35}(3x^2+8).
$$
Now apply this in annother case: $y=\ln(x^3+8x+5)$.  We have
\begin{align}
y = \ln u, & & u = x^3+8x+5 \\[10pt]
\frac{dy}{du} = \frac 1 u, & & \frac{du}{dx} = 3x^3+8
\end{align}
so
$$
\frac{dy}{dx} = \frac 1 u \cdot (3x^2+8) = \frac 1{x^3+8x+5} \cdot (3x^2+8).
$$
A: Because the rule is, for $u:\mathbb{R}\to]0,+\infty[$ a derivable function,
$$(\ln\circ u)'=\frac{u'}{u},$$ which is indeed an application of the chain rule
$$(v\circ u)'=u'\times v'\circ u$$ giving here
$$(\ln\circ u)'=u'\times\frac{1}{u}=\frac{u'}{u}.$$
In your example, $u$ is given by $u : x\mapsto5\times x$ and you derivative is
$$(ln(5x))'=\frac{5}{5\times x}=\frac{1}{x}$$ for $x\in]0,+\infty[.$
