Proofs of Derivative Properties with Examples

Here we will prove various properties of derivatives with applications one by one.

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 $\frac{d}{dx}(\sec x)$

$\frac{d}{dx}(\sec x)$

$=\frac{d}{dx}(\frac{1}{\cos x})$

$=\frac{\cos x \cdot \frac{d}{dx}(1)-1 \cdot \frac{d}{dx}(\cos x)}{\cos^2 x}$

$=\frac{\cos x \cdot 0+1 \cdot \sin x}{\cos^2 x}$

$=\frac{\sin x}{\cos^2 x}$

$=\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$

$=\sec x \tan x$

 

 

Leave a Comment