TITLE: Why do you need to use the chain rule in differentiation of ln?
QUESTION [12 upvotes]: I understand application of chain rule in the differentiation of a random function $(x^2+3)^3$. 
But, why do you need to use chain-rule when differentiating something like $\ln(2x-1)$; why won't it just be $\displaystyle\frac 1{2x-1}$? Please help.

REPLY [2 votes]: When the function of $ln$ is a composite function, like $ln (2x^2+1)$, $ln(x^2+3)$,$ln(3x^2+x)....,$ then we need the chain rule to differentiate it.
Suppose, we need to find the derivative of $ln (2x^2+1)$
Composite Function $f (g(x)) or (fog) (x)$:
When $f$ is a function of another function $g(x)$, it is called composite function.
That means $ln (2x^2+1)$ is a composite function because $ln$ is a function of another function $2x^2+1$
$f(g(x)) = ln (2x^2+1)$
Let $t(x) = 2x^2+1$ then $f(t) = ln (t)$
Reason for, Why we do not stop the differentiation of $ln (g(x))$ at just $\dfrac {1}{g(x)}$:
$\dfrac {d}{dx} \ell n(2x^2+1)$ should be equal to $\dfrac {1}{2x^2+1}$.
But $2x^2+1$ is also a function of $x$, so we can't stop it at $\dfrac {1}{2x^2+1}$
It can further be differentiated.
So, using chain rule formula, i.e.,
$\dfrac {d}{dx} f (g(x)) = f' (g(x) × g'(x)$
$\dfrac {d}{dx} \ell n (2x^2+1)=\dfrac {1}{2x^2+1}×\dfrac {d}{dx}(2x^2+1)$
$=\dfrac {1}{2x^2+1}×4x$
Thus, derivative of $\ell n (2x^2+1)$ is $\dfrac {4x}{2x^2+1}$
Chain Rule Formula:
We use the chain rule formula to differentiate the composite functions $f(g(x)),$
$\dfrac {d}{dx} f(g(x)) = f' (g(x)) × g'(x)$ ... (i)

Example: Find the derivative of $e^{(x^2+1)}$
So, $f(g(x)) = e^{(x^2+1)}$
$\therefore f'(g(x))=e^{(x^2+1)}$
and  $g(x) = x^2+1$
$\therefore g' (x) = 2x $
Put all these values in equation (i)
$\dfrac {d}{dx}e^{(x^2+1)}=e^{(x^2+1)}×2x$
Thus, 
$ \dfrac {d}{dx} e^{(x^2+1)}= 2x \cdot e^{(x^2+1)}$