Proofs of Derivative Formulas by Definition:
In this section, we will evaluate derivatives of functions u\sing the first principle. For example, we calculate the derivatives of $x^n$, $e^x$, $\sin x$, $\log x$ etc. The following formulas will be proved by the definition of derivatives:
1. $\frac{d}{dx}$(xn)=nxn-1
2. $\frac{d}{dx}$(ex)=ex
3. $\frac{d}{dx}$(ax) = ax logea
4. $\frac{d}{dx}$(log x) = 1/x (x ≠ 0)
5. $\frac{d}{dx}$(sin x) = cos x
6. $\frac{d}{dx}$(cos x) = -sin x
7. $\frac{d}{dx}$(tan x) = sec2x
At first, we calculate the derivative of xn by the definition of the derivative.
Problem 1: $\frac{d}{dx}(x^n)=nx^{n-1}$
Solution:
Let y=f(x)=xn.
From the definition of derivatives, we get that
\frac{dy}{dx} = limh→0 $\frac{f(x+h)-f(x)}{h}$ = limh→0 $\frac{(x+h)^n-x^n}{h}$
Assume z=x+h. Then z→x as h→0. Hence
\frac{dy}{dx} = limz→x $\frac{z^n-x^n}{z-x}$ [∵ h=z-x]
= nxn-1 [∵ limx→a(xn -an)/(x-a)=nan-1]
Therefore, $\frac{d}{dx}(x^n)=nx^{n-1}$.
Next, we calculate the derivative of $e^x$ by the definition of the derivative.
Problem 2: $\frac{d}{dx}(e^x)=e^x$
Solution:
Let $y=f(x)=e^x$
From the definition of derivatives, we get that
\frac{dy}{dx} = limh→0 $\frac{f(x+h)-f(x)}{h}$ = limh→0 $\frac{e^{x+h}-e^x}{h}$
= limh→0 $\frac{e^x cdot e^h – e^x}{h}$ $=e^x \cdot \lim_{h to 0} \frac{e^h-1}{h}$
$=e^x \cdot 1$ [∵ limx→0(ex -1)/x = 1]
= ex
Therefore, $\frac{d}{dx}$(ex) = ex.
Next, we calculate the derivative of ax by the definition of the derivative.
Problem 3: $\frac{d}{dx}(a^x)=a^x \log_e a$
Solution:
Let y=f(x)=ax.
From the definition of derivatives, we get that
$\frac{dy}{dx}$ = limh→0 $\frac{f(x+h)-f(x)}{h}$ = limh→0 $\frac{a^{x+h}-a^x}{h}$
= limh→0 $\frac{a^x cdot a^h – a^x}{h}$
$=a^x \cdot \lim_{h to 0} \frac{a^h-1}{h}$
= ax logea (a>0) [∵ limx→0(ax -1)/x = logea]
Therefore, $\frac{d}{dx}$ (ax) = ax logea provided that a>0.
Next, we calculate the derivative of logx by the definition of the derivative.
Problem 4: $\frac{d}{dx}(\log x)=1/x$
Solution:
Let y=f(x)=log x.
From the definition of derivatives, we get that
\frac{dy}{dx} = lim_{h to 0} \frac{f(x+h)-f(x)}{h}$ $=lim_{h to 0} \frac{\log(x+h)-\log x}{h}$
$=lim_{h to 0} \frac{1}{h} \log(\frac{x+h}{x})$ [$because \log a – \log b=\log a/b$]
$=lim_{h to 0} \frac{1}{h} \log(1+h/x)$
$=lim_{h to 0} \frac{\log(1+h/x)}{h/x} cdot 1/x$
[Put $z=h/x$. Then $z to 0$ as $h to 0$]
$=1/xlim_{z to 0} \frac{\log(1+z)}{z}$
$=1/x cdot 1$ [$because lim_{x to 0} \frac{\log(1+x)}{x}=1$]
$=1/x (x ne 0)$
Therefore, $\frac{d}{dx}(\log x)=1/x (x ne 0)$.
Next, we calculate the derivative of \sin x by the definition of the derivative.
Problem 5: $\frac{d}{dx}(\sin x)=\cos x$
Solution:
Let $y=f(x)=\sin x$
From the definition of derivatives, we get that
$\frac{dy}{dx}=lim_{h to 0} \frac{f(x+h)-f(x)}{h}$ $=lim_{h to 0} \frac{\sin(x+h)-\sin x}{h}$
[Formula used: $ \sin a -\sin b=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}$]
= limh→0 $\frac{1}{h} cdot 2 \cos \frac{2x+h}{2}\sin \frac{h}{2}$
= limh→0 $\cos(x+h/2) cdot lim_{h \to 0} \frac{\sin h/2}{h/2}$
[Put z=h/2. Then $z \to 0$ as $h \to 0$]
= limh→0 $\cos(x+h/2) \cdot \lim_{z \to 0} \frac{\sin z}{z}$
= cos(x+0) ⋅ 1 [∵ limx→0 sinx/x = 1]
= cos x.
So the derivative of sinx by limit definition is cosx.
Next, we calculate the derivative of \cos x by the definition of the derivative.
Problem 6: $\frac{d}{dx}(\cos x)=-\sin x$
Solution:
Let $y=f(x)=cos x.
From the definition of derivatives, we get that
$\frac{dy}{dx}$ = limh→0 $\frac{f(x+h)-f(x)}{h}$ = limh→0 $\frac{\cos(x+h)-\cos x}{h}$
[Formula used: cosa -cosb = 2 $\sin \frac{a+b}{2}\sin \frac{b-a}{2}$]
= limh→0 $\frac{1}{h} \cdot 2 \sin \frac{2x+h}{2}\sin \frac{-h}{2}$
= – limh→0 sin(x+h/2) ⋅ limh→0 $\frac{\sin h/2}{h/2}$ [∵ sin(-x) = -sin x]
[Put z=h/2. Then z→0 as h→0]
= – limh→0 sin(x+h/2) ⋅ limz→0 $\frac{\sin z}{z}$
= – sin(x+0) ⋅ 1 [∵ limx→0 sinx/x = 1]
= – sin x
Therefore, the derivative of cosx by limit definition is -sinx.
Next, we calculate the derivative of tan x by the definition of the derivative.
Problem 7: $\frac{d}{dx}(tan x)=sec^2 x$
Solution:
Let y=f(x)=tan x
From the definition of derivatives, we get that
$\frac{dy}{dx}$ = limh→0 $\frac{f(x+h)-f(x)}{h}$ = limh→0 $\frac{\tan(x+h)-\tan x}{h}$
= limh→0 $\frac{1}{h} [\frac{\sin(x+h)}{\cos(x+h)}-\frac{\sin x}{\cos x}]$
= limh→0 $\frac{1}{h} [\frac{\sin(x+h)\cos x – \cos(x+h) \sin x}{\cos(x+h)\cos x}]$
= limh→0 $\frac{1}{h} [\frac{\sin(x+h-x)}{\cos(x+h) \cos x}]$
= limh→0 $\frac{\sin h}{h}$ ⋅ limh→0 $\frac{1}{\cos(x+h) \cos x}$
[Formula used: sina cosb -cosa sinb = sin(a-b)]
= 1 ⋅ $\frac{1}{\cos x \cos x}$ [∵ limx→0 sinx/x = 1]
= sec2x [∵ secx = 1/cosx]
Therefore, the derivative of tanx by limit definition is sec2x.