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)=nx^n-1

2. $\frac{d}{dx}$(e^x)=e^x

3. $\frac{d}{dx}$(a^x) = a^x log_ea

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)=x^n.

From the definition of derivatives, we get that

\frac{dy}{dx} = lim_h→0 $\frac{f(x+h)-f(x)}{h}$ = lim_h→0 $\frac{(x+h)^n-x^n}{h}$

Assume z=x+h. Then z→x as h→0. Hence

\frac{dy}{dx} = lim_z→x $\frac{z^n-x^n}{z-x}$  [∵ h=z-x]

= nx^n-1 [∵ lim_x→a(xⁿ -aⁿ)/(x-a)=na^n-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} = lim_h→0 $\frac{f(x+h)-f(x)}{h}$ = lim_h→0 $\frac{e^{x+h}-e^x}{h}$

= lim_h→0 $\frac{e^x cdot e^h – e^x}{h}$ $=e^x \cdot \lim_{h to 0} \frac{e^h-1}{h}$

$=e^x \cdot 1$ [∵ lim_x→0(e^x -1)/x = 1]

= e^x

Therefore, $\frac{d}{dx}$(e^x) = e^x.

 

Next, we calculate the derivative of a^x by the definition of the derivative.

Problem 3:   $\frac{d}{dx}(a^x)=a^x \log_e a$

Solution:

Let y=f(x)=a^x.

From the definition of derivatives, we get that

$\frac{dy}{dx}$ = lim_h→0 $\frac{f(x+h)-f(x)}{h}$ = lim_h→0 $\frac{a^{x+h}-a^x}{h}$

= lim_h→0 $\frac{a^x cdot a^h – a^x}{h}$

$=a^x \cdot \lim_{h to 0} \frac{a^h-1}{h}$

= a^x log_ea (a>0) [∵ lim_x→0(a^x -1)/x = log_ea]

Therefore, $\frac{d}{dx}$ (a^x) = a^x log_ea 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}$]

= lim_h→0 $\frac{1}{h} cdot 2 \cos \frac{2x+h}{2}\sin \frac{h}{2}$

= lim_h→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 $h \to 0$]

= lim_h→0 $\cos(x+h/2) \cdot \lim_{z \to 0} \frac{\sin z}{z}$

= cos(x+0) ⋅ 1 [∵ lim_x→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

$\frac{dy}{dx}$ = lim_h→0 $\frac{f(x+h)-f(x)}{h}$ = lim_h→0 $\frac{\cos(x+h)-\cos x}{h}$

[Formula used: cosa -cosb = 2 $\sin \frac{a+b}{2}\sin \frac{b-a}{2}$]

= lim_h→0 $\frac{1}{h} \cdot 2 \sin \frac{2x+h}{2}\sin \frac{-h}{2}$

= – lim_h→0 sin(x+h/2) ⋅ lim_h→0 $\frac{\sin h/2}{h/2}$ [∵ sin(-x) = -sin x]

[Put z=h/2. Then z→0 as h→0]

= – lim_h→0 sin(x+h/2) ⋅ lim_z→0 $\frac{\sin z}{z}$

= – sin(x+0) ⋅ 1  [∵ lim_x→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

$\frac{dy}{dx}$ = lim_h→0 $\frac{f(x+h)-f(x)}{h}$ = lim_h→0 $\frac{\tan(x+h)-\tan x}{h}$

= lim_h→0 $\frac{1}{h} [\frac{\sin(x+h)}{\cos(x+h)}-\frac{\sin x}{\cos x}]$

= lim_h→0 $\frac{1}{h} [\frac{\sin(x+h)\cos x – \cos(x+h) \sin x}{\cos(x+h)\cos x}]$

= lim_h→0 $\frac{1}{h} [\frac{\sin(x+h-x)}{\cos(x+h) \cos x}]$

= lim_h→0 $\frac{\sin h}{h}$ ⋅ lim_h→0 $\frac{1}{\cos(x+h) \cos x}$

[Formula used: sina cosb -cosa sinb = sin(a-b)]

= 1 ⋅ $\frac{1}{\cos x \cos x}$ [∵ lim_x→0 sinx/x = 1]

= sec²x [∵ secx = 1/cosx]

Therefore, the derivative of tanx by limit definition is sec²x.