Here we will prove various properties of derivatives with applications one by one.
Table of Contents
Derivative of a constant function is zero- proof:
For a constant $c$, we have
$\frac{d}{dx}(c)=0$
Proof:
Let $f(x)=c$
Now, $\frac{d}{dx}(c)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}{ \large \frac{f(x+h)-f(x)}{h} }$
$=\lim\limits_{h \to 0}{ \large \frac{c-c}{h} }$
$=\lim\limits_{h \to 0}{ \large \frac{0}{h} }$
$=\lim\limits_{h \to 0}0$
$=0$ (proved)
Proof of the derivative of a scalar multiple:
For a constant $c$ and a differentiable function $f(x)$, we have
$\frac{d}{dx}[c \cdot f(x)]=c \cdot \frac{d}{dx}(f(x))$
Proof:
Let $u(x)=c \cdot f(x)$
The definition of the derivative implies that
$\frac{d}{dx}[c \cdot f(x)]$
$=\frac{d}{dx}(u(x))$ $=\lim\limits_{h \to 0}\frac{u(x+h)-u(x)}{h}$
$=\lim\limits_{h \to 0}\frac{cf(x+h)-cf(x)}{h}$
$=\lim\limits_{h \to 0}\frac{c[f(x+h)-f(x)]}{h}$
$=c\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$=c \frac{d}{dx}(f(x))$ (proved)
Proof of the sum rule of derivative:
For two differentiable functions $f(x)$ and $g(x)$, we have
$\frac{d}{dx}[f(x)+g(x)]$ $=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$
Proof:
Let $u(x)=f(x)+g(x)$
Now, $\frac{d}{dx}[f(x)+g(x)]$
$=\frac{d}{dx}(u(x))$
$=\lim\limits_{h \to 0} { \large \frac{u(x+h)-u(x)}{h}}$
$=\lim\limits_{h \to 0} { \large\frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h} }$
$=\lim\limits_{h \to 0} { \large \frac{[f(x+h)-f(x)]+[g(x+h)-g(x)]}{h} }$
$=\lim\limits_{h \to 0} { \large [\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}] }$
$=\lim\limits_{h \to 0}{ \large \frac{f(x+h)-f(x)}{h} }$ $+\lim\limits_{h \to 0} { \large \frac{g(x+h)-g(x)}{h} }$
$=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$ (proved)
Replacing the plus sign with minus sign in the above proof, we can obtain the difference rule of derivative:
$\frac{d}{dx}[f(x)-g(x)]$ $=\frac{d}{dx}(f(x))-\frac{d}{dx}(g(x)).$
Proof of the product rule of derivative:
For two differentiable functions $f(x)$ and $g(x)$, we have
$\frac{d}{dx}[f(x)g(x)]$ $=f(x)\frac{d}{dx}(g(x))+g(x)\frac{d}{dx}(f(x))$
Proof:
Let $u(x)=f(x)g(x)$
Thus, ${ \large\frac{u(x+h)-u(x)}{h}}$
$={ \large\frac{f(x+h)g(x+h)-f(x)g(x)}{h}}$
$={ \large\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}}$
$={ \large\frac{f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)]}{h}}$
$\therefore \frac{d}{dx}[f(x)g(x)]$
$=\frac{d}{dx}(u(x))$
$=\lim\limits_{h \to 0}\frac{u(x+h)-u(x)}{h}$
$=\lim\limits_{h \to 0}$ $\frac{f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)]}{h}$
$=\lim\limits_{h \to 0}[f(x+h)\frac{g(x+h)-g(x)}{h}$ $+g(x)\frac{f(x+h)-f(x)}{h}]$
$=\lim\limits_{h \to 0}f(x+h)\frac{g(x+h)-g(x)}{h}$ $+\lim\limits_{h \to 0}g(x)\frac{f(x+h)-f(x)}{h}$
$=f(x+0)\frac{d}{dx}(g(x))+g(x)\frac{d}{dx}(f(x))$
$=f(x)\frac{d}{dx}(g(x))+g(x)\frac{d}{dx}(f(x))$ (proved)
Proof of the quotient rule of derivative:
For two differentiable functions $f(x)$ and $g(x)$, we have
$\frac{d}{dx}[\frac{f(x)}{g(x)}]$ $=\frac{g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x))}{(g(x))^2}$
Proof:
Let $u(x)=\frac{f(x)}{g(x)}$
Thus, $u(x+h)-u(x)$
$=\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}$
$=\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}$
$=\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}$
$=\frac{g(x)\{f(x+h)-f(x)\}-f(x)\{g(x+h)-g(x)\}}{g(x+h)g(x)}$
$\therefore \frac{d}{dx}[\frac{f(x)}{g(x)}]$
$=\frac{d}{dx}(u(x))$
$=\lim\limits_{h \to 0}\frac{u(x+h)-u(x)}{h}$
$=\lim\limits_{h \to 0}$ $\frac{1}{h}[\frac{g(x)\{f(x+h)-f(x)\}-f(x)\{g(x+h)-g(x)\}}{g(x+h)g(x)}]$
$=\lim\limits_{h \to 0}\frac{1}{g(x+h)g(x)}$ $[g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h}]$
$=\frac{1}{g(x)g(x)}[g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x))]$
$=\frac{g(x)\frac{d}{dx}(f(x))-f(x)\frac{d}{dx}(g(x))}{(g(x))^2}$ (proved)
Examples: (i) Find $\frac{d}{dx}(\tan x)$
$\frac{d}{dx}(\tan x)$
$=\frac{d}{dx}(\frac{\sin x}{\cos x})$
$=\frac{\cos x \frac{d}{dx}(\sin x)-\sin x \frac{d}{dx}(\cos x)}{\cos^2 x}$
$=\frac{\cos x \cdot \cos x-\sin x \cdot (-\sin x)}{\cos^2 x}$
$=\frac{\cos^2 x +\sin^2 x}{\cos^2 x}$
$=\frac{1}{\cos^2 x}$ $[\because \sin^2 x+\cos^2 x=1]$
$=\sec^2 x$
(ii) Find $\frac{d}{dx}(\sec x)$
$\frac{d}{dx}(\sec x)$
$=\frac{d}{dx}(\frac{1}{\cos x})$
$=\frac{\cos x \cdot \frac{d}{dx}(1)-1 \cdot \frac{d}{dx}(\cos x)}{\cos^2 x}$
$=\frac{\cos x \cdot 0+1 \cdot \sin x}{\cos^2 x}$
$=\frac{\sin x}{\cos^2 x}$
$=\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$
$=\sec x \tan x$
This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.