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 \[y+\Delta y=a^{x+\Delta x}\] \[\Delta y=a^{x+\Delta x}-y\] \[\Delta y=a^{x+\Delta x}-a^x\] \[\Delta y=a^x(a^{\Delta x}-1)\] \[\frac{\Delta y}{\Delta x}=\frac{a^x(a^{\Delta x}-1)}{\Delta x}\] \[\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{a^x(a^{\Delta x}-1)}{\Delta x}\] \[=a^x\lim_{\Delta x\to 0}\frac{a^{\Delta x}-1}{\Delta x}\] \[=a^x.\log a\]