Logarithmic Differentiation & Key Derivatives

Logarithmic Differentiation

• Complicated target function: $y=\frac{(1+x^2)^5(1+x^3)^8}{(1+x^4)^7}$
• Take natural log: $\ln y = 5\ln(1+x^2)+8\ln(1+x^3)-7\ln(1+x^4)$
• Differentiate implicitly: $\frac{1}{y}\,\frac{dy}{dx}=\frac{10x}{1+x^2}+\frac{24x^2}{1+x^3}-\frac{28x^3}{1+x^4}$
• Multiply by $y$ to solve: $\displaystyle \frac{dy}{dx}= y\left[\frac{10x}{1+x^2}+\frac{24x^2}{1+x^3}-\frac{28x^3}{1+x^4}\right]$ (replace $y$ with original expression to finish).

Derivative of $\ln x$

• Definition: $\displaystyle \frac{d}{dx}\ln x = \lim_{\Delta x\to 0}\frac{\ln(x+\Delta x)-\ln x}{\Delta x}$
• Combine logs: $=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\,\ln!\left(1+\frac{\Delta x}{x}\right)$
• Substitute $\tfrac{\Delta x}{x}=\tfrac{1}{n} \;(n\to\infty)$ ⇒ $\Delta x=\tfrac{x}{n}$
• Expression becomes $\frac{1}{x}\,\lim_{n\to\infty}\ln!\left(1+\frac{1}{n}\right)^{n}$
• Limit inside log equals $e$, so total limit = $\tfrac{1}{x}\cdot1$.
• Result: $\boxed{\dfrac{d}{dx}\ln x=\tfrac{1}{x}}$.

Derivative of $e^{x}$ (non-circular)

• Evaluate $\ln(e^{x})$ two ways:
  1. Simplify first: $\ln(e^{x})=x \;\Rightarrow\; \dfrac{d}{dx}=1$
  2. Chain rule: $\dfrac{d}{dx}\ln(e^{x})=\dfrac{d}{dx}(e^{x})\,\cdot\dfrac{1}{e^{x}}$
• Equate: $\dfrac{d}{dx}(e^{x})\cdot\dfrac{1}{e^{x}}=1$ ⇒ $\boxed{\dfrac{d}{dx}(e^{x})=e^{x}}$.

Key Points to Remember

• Logarithmic differentiation converts powers → coefficients and products/quotients → sums/differences, simplifying derivatives of messy functions.
• Fundamental results: $\dfrac{d}{dx}\ln x = \tfrac{1}{x}$ and $\dfrac{d}{dx}e^{x}=e^{x}$ are proved without circular reasoning.
• Use these identities to streamline differentiation of exponentials, products, and quotients.