# 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.