Here we will calculate the derivatives of some well-known functions from the first principle. For example, we will find the derivatives of the polynomial functions, trigonometric functions, exponential functions, logarithmic functions, and so on.
Firstly, we find the derivative of xn using the definition of the derivative.
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Power rule of Derivative using First Principle:
| \[\frac{d}{dx}(x^n)=nx^{n-1}\] |
Proof:
Let $f(x)=x^n.$ From the first principle of derivative, we have
$\frac{d}{dx}(x^n)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\dfrac{(x+h)^n-x^n}{h}$
[Let z=x+h. Then z → x as h → 0]
$=\lim\limits_{z \to x}\dfrac{z^n-x^n}{z-x}$ $[\because h=z-x]$
$=nx^{n-1}$ $[\because \lim\limits_{x \to a}\frac{x^n-a^n}{x-a}=nx^{n-1}]$ ♣
Next, we will find the derivative of $e^x$ by the definition of derivative.
Proof of the Derivative of ex from First Principle:
| \[\frac{d}{dx}(e^x)=e^x\] |
Proof:
Let $f(x)=e^x.$ By the definition of the derivative, we have
$\frac{d}{dx}(e^x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\dfrac{e^{x+h}-e^x}{h}$
$=\lim\limits_{h \to 0}\dfrac{e^x\cdot e^h-e^x}{h}$
$=\lim\limits_{h \to 0}\dfrac{e^x(e^h-1)}{h}$
$=e^x\lim\limits_{h \to 0}\dfrac{e^h-1}{h}$
$=e^x \cdot 1$ $[\because \lim\limits_{x \to 0}\frac{e^x-1}{x}=1]$
= ex ♣
Now, we calculate the derivative of ax by definition.
Proof of the Derivative of ax from First Principle:
| \[\frac{d}{dx}(a^x)=a^x\log_e a\] |
Proof:
Let $f(x)=a^x.$ By the definition of the derivative, we have
$\frac{d}{dx}(a^x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\dfrac{a^{x+h}-a^x}{h}$
$=\lim\limits_{h \to 0}\dfrac{a^x\cdot a^h-a^x}{h}$
$=\lim\limits_{h \to 0}\dfrac{a^x(a^h-1)}{h}$
$=a^x\lim\limits_{h \to 0}\frac{a^h-1}{h}$
$=a^x \cdot \log_e a$ $[\because \lim\limits_{x \to 0}\frac{a^x-1}{x}=\log_e a]$
$=a^x\log_e a$ ♣
Next, we find the derivative of log x by the definition of the derivative.
Proof of the Derivative of log x from First Principle:
| \[\frac{d}{dx}(\log x)=\frac{1}{x}\] |
Proof:
Let $f(x)=\log x.$ By the definition of the derivative, we have
$\frac{d}{dx}(\log x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\dfrac{\log(x+h)-\log x}{h}$
$=\lim\limits_{h \to 0}\dfrac{\log \frac{x+h}{x}}{h}$ $[\because \log a -\log b=\log \frac{a}{b}]$
$=\lim\limits_{h \to 0}\dfrac{\log (1+\frac{h}{x})}{h}$
$=\lim\limits_{h \to 0}\dfrac{1}{x}\dfrac{\log (1+\frac{h}{x})}{h/x}$
$=\dfrac{1}{x}\lim\limits_{h \to 0}\dfrac{\log (1+\frac{h}{x})}{h/x}$
[Let h/x=t. Then t → 0 as h → 0]
$=\dfrac{1}{x}\lim\limits_{t \to 0}\dfrac{\log (1+t)}{t}$
$=\dfrac{1}{x} \cdot 1$ $[\because \lim\limits_{x \to 0}\dfrac{\log(1+x)}{x}=1]$
$=\dfrac{1}{x}$ ♣
Now, we find the derivative of sin x by the definition of the derivative.
Proof of the Derivative of sin x from First Principle:
| \[\frac{d}{dx}(\sin x)=\cos x\] |
Proof:
Let $f(x)=\sin x.$ By the definition of the derivative, we have
$\frac{d}{dx}(\sin x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h}$
$=\lim\limits_{h \to 0}\dfrac{2\cos {\frac{2x+h}{2}} \sin {\frac{h}{2}}}{h}$ $[\because \sin a -\sin b$ $=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}]$
$=\lim\limits_{h \to 0}\cos \big(\dfrac{2x+h}{2}\big) \dfrac{\sin \frac{h}{2}}{h/2}$
$=\lim\limits_{h \to 0}\cos \big(\dfrac{2x+h}{2}\big) \lim\limits_{h \to 0}\dfrac{\sin \frac{h}{2}}{h/2}$
[Let h/2=t. Then t → 0 as h → 0]
$=\cos \big(\dfrac{2x+0}{2}\big) \lim\limits_{t \to 0}\dfrac{\sin t}{t}$
= cos x ⋅ 1
= cos x ♣
Let us now find the derivative of cos x by the definition of the derivative.
Proof of the Derivative of cos x from First Principle:
| \[\frac{d}{dx}(\cos x)=-\sin x\] |
Proof:
Let $f(x)=\cos x.$ By the definition of the derivative, we have
$\frac{d}{dx}(\cos x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h}$
$=\lim\limits_{h \to 0} \dfrac{2\sin {\frac{2x+h}{2}} \sin {\frac{-x}{2}}}{h}$ $[\because \cos a -\cos b$ $=2\sin \frac{a+b}{2}\sin \frac{b-a}{2}]$
$=-\lim\limits_{h \to 0}\sin (\frac{2x+h}{2}) \frac{\sin \frac{h}{2}}{h/2}$ $[\because \sin(-x)=-\sin x]$
$=-\lim\limits_{h \to 0}\sin (\frac{2x+h}{2}) \lim\limits_{h \to 0}\frac{\sin \frac{h}{2}}{h/2}$
[Let h/2=t. Then t → 0 as h → 0]
$=-\sin (\frac{2x+0}{2}) \lim\limits_{t \to 0}\frac{\sin t}{t}$
= -sin x ⋅ 1
= -sin x ♣
We now find the derivative of tan x by the definition of the derivative.
Proof of the Derivative of tan x from First Principle:
| \[\frac{d}{dx}(\tan x)=\sec^2 x\] |
Proof:
Let $f(x)=\tan x.$
Now, $\dfrac{f(x+h)-f(x)}{h}$
$=\dfrac{\tan(x+h)-\tan x}{h}$
$=\dfrac{1}{h}\Big[\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}\Big]$
$=\frac{1}{h}\big[\frac{\sin(x+h)\cos x -\sin x\cos(x+h)}{\cos (x+h)\cos x}\big]$
$=\dfrac{1}{h}\Big[\dfrac{\sin(x+h-x)}{\cos (x+h)\cos x}\Big]$
$[\because \sin (a-b)$ $=\sin a \cos b -\cos a \sin b]$
$=\dfrac{1}{h}\Big[\dfrac{\sin h}{\cos (x+h)\cos x}\Big]$
So by the definition of the derivative, we have
$\frac{d}{dx}(\tan x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{1}{h} \Big[\dfrac{\sin h}{\sin (x+h)\cos x}\Big]$
$=\lim\limits_{h \to 0} \big[ {\large \frac{\sin h}{h}} {\large \frac{1}{\cos (x+h)\cos x}} \big]$
$=\lim\limits_{h \to 0} {\large \frac{\sin h}{h}}$ $\cdot \lim\limits_{h \to 0} {\large \frac{1}{\cos (x+h)\cos x}}$
$=1 \cdot \dfrac{1}{\cos(x+0)\cos x}$
$=1 \cdot \dfrac{1}{\cos^2 x}$
= sec2 x ♣
Next, we calculate the derivative of cot x by the definition of the derivative.
Proof of the Derivative of cot x from First Principle:
| \[\frac{d}{dx}(\cot x)=-\text{cosec}^2 x\] |
Proof:
Let $f(x)=\cot x.$
Now, $\dfrac{f(x+h)-f(x)}{h}$
$=\dfrac{\cot(x+h)-\cot x}{h}$
$=\dfrac{1}{h}\Big[\dfrac{\cos(x+h)}{\sin(x+h)}-\dfrac{\cos x}{\sin x}\Big]$
$=\frac{1}{h}\big[\frac{\sin x\cos(x+h) -\cos x\sin(x+h)}{\sin (x+h)\sin x}\big]$
$=\dfrac{1}{h}\Big[\dfrac{\sin(x-x-h)}{\cos (x+h)\cos x}\Big]$
$[\because \sin (a-b)=$ $\sin a \cos b -\cos a \sin b]$
$=\frac{1}{h}{\large \big[\frac{-\sin h}{\sin (x+h)\sin x}\big]}$ $[\because \sin(-x)=-\sin x]$
So by the definition of the derivative, we have
$\frac{d}{dx}(\cot x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0}\frac{1}{h}\Big[\dfrac{-\sin h}{\sin (x+h)\sin x}\Big]$
$=-\lim\limits_{h \to 0} \big[ {\large \frac{\sin h}{h}} {\large \frac{1}{\sin (x+h)\sin x}} \big]$
$=-\lim\limits_{h \to 0} {\large \frac{\sin h}{h}}$ $\cdot \lim\limits_{h \to 0} {\large \frac{1}{\sin (x+h)\sin x}}$
$=-1 \cdot \dfrac{1}{\sin(x+0)\sin x}$
$=-1 \cdot \dfrac{1}{\sin^2 x}$
$=-\text{cosec}^2 x$ ♣
Now, we evaluate the derivative of sec x by the definition of the derivative.
Proof of the Derivative of sec x From First Principle:
| \[\frac{d}{dx}(\sec x)=\sec x \tan x\] |
Proof:
Let $f(x)=\sec x.$
Now, $\dfrac{f(x+h)-f(x)}{h}$
$=\dfrac{\sec(x+h)-\sec x}{h}$
$=\dfrac{1}{h}\Big[\dfrac{1}{\cos(x+h)}-\dfrac{1}{\cos x}\Big]$
$=\dfrac{1}{h}\Big[\dfrac{\cos x -\cos(x+h)}{\cos (x+h)\cos x}\Big]$
$=\dfrac{1}{h}\Big[\dfrac{2\sin \frac{2x+h}{2}\sin \frac{h}{2}}{\cos (x+h)\cos x}\Big]$ $[\because \cos a- \cos b)$ $=2\sin \frac{a+b}{2}\sin \frac{b-a}{2}]$
$={\large \big[\frac{\sin h/2}{h/2} \cdot \frac{\sin \frac{2x+h}{2}}{\cos (x+h)\cos x}\big]}$
So by the definition of the derivative, we have
$\dfrac{d}{dx}(\sec x)$ $=\dfrac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=\lim\limits_{h \to 0} {\large \big[\frac{\sin h/2}{h/2} \cdot \frac{\sin \frac{2x+h}{2}}{\cos (x+h)\cos x}\big]}$
$=\lim\limits_{h \to 0} {\large\frac{\sin h/2}{h/2}}$ $\cdot \lim\limits_{h \to 0}{\large \frac{\sin \frac{2x+h}{2}}{\cos (x+h)\cos x}}$
$=1 \cdot \dfrac{\sin \frac{2x+0}{2}}{\cos (x+0)\cos x}$
$=\dfrac{\sin x}{\cos^2 x}$
$=\dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}$
$=\sec x \tan x$ ♣
At last, we find the derivative of cosec x by the definition of the derivative.
Proof of the Derivative of cosec x from First Principle:
| \[\frac{d}{dx}(\text{cosec}\, x)=-\text{cosec}\, x \cot x\] |
Proof:
Let $f(x)=\text{cosec}\, x.$
Now, $\dfrac{f(x+h)-f(x)}{h}$
$=\dfrac{\text{cosec}(x+h)-\text{cosec}\, x}{h}$
$=\dfrac{1}{h}\Big[\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin x}\Big]$
$=\dfrac{1}{h}\Big[\dfrac{\sin x -\sin(x+h)}{\sin (x+h)\sin x}\Big]$
$=\dfrac{1}{h}\Big[\dfrac{2\cos \frac{2x+h}{2}\sin \frac{-h}{2}}{\sin (x+h)\sin x}\Big]$ $[\because \sin a- \sin b)$ $=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}]$
$=-{\large \big[\frac{\sin h/2}{h/2} \cdot \frac{\cos \frac{2x+h}{2}}{\sin (x+h)\sin x}\big]}$
So by the definition of the derivative, we have
$\frac{d}{dx}(\text{cosec}\, x)$ $=\frac{d}{dx}(f(x))$
$=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
$=-\lim\limits_{h \to 0} \big[\frac{\sin h/2}{h/2} \cdot$ $\frac{\cos \frac{2x+h}{2}}{\sin (x+h)\sin x}\big]$
$=-\lim\limits_{h \to 0} {\large\frac{\sin h/2}{h/2}}$ $\cdot \lim\limits_{h \to 0}{\large \frac{\cos \frac{2x+h}{2}}{\sin (x+h)\sin x}}$
$=-1 \cdot \dfrac{\cos \frac{2x+0}{2}}{\sin (x+0)\sin x}$
$=-\dfrac{\cos x}{\sin^2 x}$
$=-\dfrac{1}{\sin x}\cdot \dfrac{\cos x}{\sin x}$
$=-\text{cosec}\, x \cot x$ ♣
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.