In this note, we study derivatives of logarithmic and exponential functions.

**Derivatives of Logarithmic Functions**

First recall that \begin{equation}\label{eq:euler}\lim_{t\to 0}(1+t)^{\frac{1}{t}}=e\end{equation}\begin{align*}\frac{d}{dx}\ln x&=\lim_{h\to 0}\frac{\ln(x+h)-\ln x}{h}\\&=\lim_{h\to 0}\frac{1}{h}\ln\left(\frac{x+h}{x}\right)\\&=\frac{1}{x}\lim_{h\to 0}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\\&=\frac{1}{x}\lim_{t\to 0}\ln(1+t)^{\frac{1}{t}}\\&=\frac{1}{x}\end{align*} with $t=\frac{h}{x}$.\begin{equation}\label{eq:dln}\frac{d}{dx}\ln x=\frac{1}{x}\end{equation} Using the change of base formula $\log_ax=\frac{\ln x}{\ln a}$, we obtain \begin{equation}\label{eq:dlog}\frac{d}{dx}\log_ax=\frac{1}{x\ln a}\end{equation}

**Derivatives of Exponential Functions**

We can find the derivative of the natural exponential function $y=e^x$ using the relationship $x=\ln y$ and implicit differentiation. Differentiating $x=\ln y$ with respect to $x$ we obtain $1=\frac{1}{y}\frac{dy}{dx}$ i.e. $\frac{dy}{dx}=y=e^x$. Hence \begin{equation}\label{eq:dnatexp}\frac{d}{dx}e^x=e^x\end{equation} Note that $a^x=e^{x\ln a}$. So by the chain rule we find $$\frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln a}=e^{x\ln a}\ln a=a^x\ln a$$ Hence\begin{equation}\label{label:dexp}\frac{d}{dx}a^x=a^x\ln a\end{equation}

**The Power Rule (General Form)**

Let us consider $x^n$ for any $x>0$ and any real number $n$. As we have seen above $x^n=e^{n\ln x}$ so by the chain rule $$\frac{d}{dx}x^n=\frac{d}{dx}e^{n\ln x}=e^{n\ln x}\frac{n}{x}=nx^{n-1}$$ This completes the proof of the general power rule.

**Logarithmic Differentiation**

The derivatives of functions involving products, quotients, and powers may be found more easily (quickly) by taking the natural logarithm of such functions before differentiating. This allows us to break a complicated function into simpler pieces using properties of the natural logarithm. This whole process, which is called *logarithmic differentiation*, makes differentiation much easier and quicker.

*Example*. Use logarithmic differentiation to find the derivative of $y=\frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}$.

*Solution*. \begin{align*}\ln y&=\ln \frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}\\&=\ln x+\frac{1}{2}\ln(x^2+1)-\frac{2}{3}\ln(x+1)\end{align*} Differentiating with respect to $x$, $$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}$$ Therefore, $$\frac{dy}{dx}=\left[\frac{1}{x}+\frac{x}{x^2+1}-\frac{2}{3(x+1)}\right]\frac{x\sqrt{x^2+1}}{(x+1)^{\frac{2}{3}}}$$