Calculus – Logarithmic Derivatives, Chain Rule, & Second-Derivative Examples

The natural logarithm $\ln x$ is only defined for positive inputs (domain x>0).
To extend logarithmic expressions to negative inputs we use the absolute-value form $\ln|x|$ , whose domain is $x \neq 0$ .
Piecewise definition of the absolute value that was recalled:
- $|x| = x$ if x>0
- $|x| = -x$ if x<0
Consequences for the logarithm:
- $\ln|x| = \ln x$ when x>0
- $\ln|x| = \ln(-x)$ when $x<0$ (notice that $-x>0$ so the argument of the log is positive).

Use the chain rule separately on each branch.
1. If x>0: $\dfrac{d}{dx}\,[\ln x] = \dfrac1x$ .
2. If $x<0$ : write $u=-x$ (so $u>0$ ). Then $\ln|x| = \ln u$ and
  $\dfrac{d}{dx}\,[\ln u] = \dfrac1u\,\dfrac{du}{dx}=\dfrac1{-x}\cdot(-1)=\dfrac1x.$
Result: $\boxed{\displaystyle \frac{d}{dx}\,\ln|x| = \frac1x\qquad (x\neq 0)}.$
Antiderivative connection: $\displaystyle \int \frac1x\,dx = \ln|x|+C.$

For any differentiable function f(x)>0:
$\boxed{\displaystyle \frac{d}{dx}\,[\ln f(x)] = \frac{f'(x)}{f(x)}}.$
Special cases discussed:
- $\dfrac{d}{dx}\,[\ln(-x)] = \dfrac{-1}{-x}=\dfrac1x.$
- $\dfrac{d}{dx}\,[\ln(x^3)] = \dfrac{3x^2}{x^3}=\dfrac3x.$

Product rule: $(uv)' = u'v + uv'.$
Quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}.$
Chain rule (restated): if $y=f(u)$ and $u=g(x)$ , then $\dfrac{dy}{dx}=f'(u)\,g'(x).$

Think of $y = (\ln x)\,x^{1/2}.$ Apply the product rule.
Derivatives of the factors:
- $\dfrac{d}{dx}[\ln x] = \dfrac1x.$
- $\dfrac{d}{dx}[x^{1/2}] = \dfrac{1}{2}x^{-1/2}=\dfrac1{2\sqrt{x}}.$
Product rule gives
\begin{aligned}
y' &= (\ln x)\left(\frac1{2\sqrt{x}}\right) + \sqrt{x}\left(\frac1x\right)\[4pt]
&= \frac{\ln x}{2\sqrt{x}} + \frac1{\sqrt{x}}\[4pt]
&= \frac{1 + \tfrac12\ln x}{\sqrt{x}}.
\end{aligned}

Rewrite $y'$ as $x^{-1/2}\bigl(1+\tfrac12\ln x\bigr).$ Let
- $g(x)=x^{-1/2},\; g'(x)=-\tfrac12x^{-3/2}.$
- $h(x)=1+\tfrac12\ln x,\; h'(x)=\tfrac12\,x^{-1}.$
Apply the product rule again:
\begin{aligned}
y'' &= g'(x)h(x)+g(x)h'(x)\[4pt]
&=\left(-\frac12x^{-3/2}\right)\Bigl(1+\tfrac12\ln x\Bigr)+x^{-1/2}\left(\frac12x^{-1}\right)\[6pt]
&=x^{-3/2}\Bigl[-\tfrac12!\left(1+\tfrac12\ln x\right)+\tfrac12\Bigr]\[6pt]
&=x^{-3/2}\Bigl[-\tfrac14\ln x\Bigr]\[4pt]
&=\boxed{-\dfrac{\ln x}{4x^{3/2}}}.
\end{aligned}

Let $u = 1+\ln x.$
- $y = \ln u,\quad u' = \dfrac1x.$
Chain rule: $y' = \dfrac1u\,u' = \frac1{1+\ln x}\cdot\frac1x = \boxed{\dfrac1{x\bigl(1+\ln x\bigr)}}.$

Write $y' = \bigl[x\,(1+\ln x)\bigr]^{-1}.$
Differentiate as $y'' = -\dfrac{g'(x)}{g(x)^2},$ where $g(x)=x\,(1+\ln x).$
Compute $g'(x):$
$g'(x)=1\cdot(1+\ln x)+x\cdot\frac1x = (1+\ln x)+1 = \ln x+2.$
Therefore
$\boxed{\displaystyle y'' = -\frac{\ln x + 2}{x^{2}\,(1+\ln x)^{2}}}.$

Absolute value inside a log allows differentiation across both positive and negative inputs while excluding $x=0$ .
The derivative $\dfrac1x$ appears repeatedly; it is both the slope of $\ln|x|$ and the “inside derivative” whenever $\ln x$ is nested in larger functions.
The chain rule, product rule, and quotient rule knit together seamlessly when higher-order derivatives are required.
Integration of $\dfrac1x$ naturally yields $\ln|x|$ , illustrating the reciprocal relationship between differentiation and antidifferentiation for logarithmic functions.

Always state the domain before differentiating logarithmic expressions.
When you see $\ln(f(x))$ , immediately think “ $f'(x)/f(x)$ ,” then decide if further chain-rule steps are needed.
For higher derivatives, rewrite expressions with powers (e.g.
$\sqrt{x}=x^{1/2}$ ) so exponents can be handled mechanically.
A clean algebraic simplification at each step prevents sign errors—especially important when negative signs originate from chain‐rule substitutions like $u=-x$ .