# Derivatives of Logarithmic and Exponential Functions

Here, we shall find the derivatives of logarithmic and exponential functions. Any function of the form $ƒ(x)=a^x$ is known as exponential function, in which the base $a$ is a constant and the index $x$ is a variable. The inverse of an exponential function is called a logarithmic function which is denoted by $\log_ax$. So if $y=a^x$, then $\log_ay=x$

There is a special type of exponential function $e^x$, where $e$ is the limiting value of $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$

The value of $e$ lies between $2$ and $3$ and is approximately $2.718$. The corresponding logarithmic function is known as the natural logarithmic function and is denoted by $\log x$, base $e$ being understood.

Hence, we have, $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$

Put $n=\frac{1}{h}\Rightarrow h=\frac{1}{n}$ so that when $n\to\infty$, $h\to 0$. $\therefore e=\lim_{h\to 0}(1+h)^{\frac{1}{h}}$

Now, we can find the derivatives of the exponential and the logarithmic functions by first principles using the following limit theorems: $\text{1. }\lim_{x\to 0}\frac{\log(1+x)}{x}=1$ $\text{2. }\lim_{x\to 0}\frac{e^x-1}{x}=1$ $\text{3. }\lim_{x\to 0}\frac{a^x-1}{x}=\log a$

## Derivative of Natural Logarithmic Function $(\log x)$

Let $y=\log x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then $y+\Delta y=\log(x+\Delta x)$ $\Delta y=\log(x+\Delta x)-y$ $\Delta y=\log(x+\Delta x)-\log x$ $\frac{\Delta y}{\Delta x}=\frac{\log (x+\Delta x)-\log x}{\Delta x}$ $=\frac{\log\left(\frac{x+\Delta x}{x}\right)}{\Delta x}$ $=\frac{1}{\Delta x}\log\left(1+\frac{\Delta x}{x}\right)$ $\therefore \frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\log\left(1+\frac{\Delta x}{x}\right)$ $=\frac{1}{x}\lim_{\Delta x\to 0}\frac{\log\left(1+\frac{\Delta x}{x}\right)}{\frac{\Delta x}{x}}$ $=\frac{1}{x}.1=\frac{1}{x}$

## Derivative of $e^x$

Let $y=e^x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$.

Then $y+\Delta y=e^{x+\Delta x}$ $\Delta y=e^{x+\Delta x}-y$ $\Delta y=e^{x+Delta x}-e^x$ $\Delta y=e^x(e^{\Delta x}-1)$ $\frac{\Delta y}{\Delta x}=\frac{e^x(e^{\Delta x}-1)}{\Delta x}$ Now, $\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$ $=\lim_{\Delta x\to 0}\frac{e^x(e^{\Delta x}-1)}{\Delta x}$ $=e^x\lim_{\Delta x\to 0}\frac{e^{\Delta x}-1}{\Delta x}$ $=e^x.1=e^x$

## Derivative of Logarithmic Functions $(\log_ax)$

Let $y=\log_ax$ Changing the base from $a$ to $e$, $y=\log_ae.\log x$ $\therefore \frac{dy}{dx}=\log_ae.\frac{d(\log x)}{dx}$ $=\log_ae.\frac{1}{x}$ $=\frac{1}{x}\log_ae$

## Derivative of $a^x$

Let $y=a^x$. Let $\Delta x$ be a small increment in $x$ and $\Delta y$ be the corresponding increment in $y$. Then $y+\Delta y=a^{x+\Delta x}$ $\Delta y=a^{x+\Delta x}-y$ $\Delta y=a^{x+\Delta x}-a^x$ $\Delta y=a^x(a^{\Delta x}-1)$ $\frac{\Delta y}{\Delta x}=\frac{a^x(a^{\Delta x}-1)}{\Delta x}$ $\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{a^x(a^{\Delta x}-1)}{\Delta x}$ $=a^x\lim_{\Delta x\to 0}\frac{a^{\Delta x}-1}{\Delta x}$ $=a^x.\log a$