Proofs of Derivative Formulas

Proofs of Derivative Formulas by Definition: 

In this section, we will evaluate derivatives of functions using 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. `d/dx(x^n)=nx^{n-1}`

2. `d/dx(e^x)=e^x`

3. `d/dx(e^x)=a^x log_e a`

4. `d/dx(log x)=1/x (x ne 0)`

5. `d/dx(sin x)=cos x`

6. `d/dx(cos x)=-sin x`

7. `d/dx(tan x)=sec^2 x`

At first, we calculate the derivative of  `x^n` by the definition of the derivative.

Problem 1:    `d/dx(x^n)=nx^{n-1}`

Solution:

Let `y=f(x)=x^n`

From the definition of derivatives, we get that 

`dy/dx=lim_{h to 0} frac{f(x+h)-f(x)}{h}` `=lim_{h to 0} frac{(x+h)^n-x^n}{h}`

Assume `z=x+h.` Then `z to x` as `h to 0`. Hence

`dy/dx=lim_{z to x} frac{z^n-x^n}{z-x}`  [`because h=z-x`]

`=nx^{n-1}` 

[`because lim_{x to a}frac{x^n-a^n}{x-a}=na^{n-1}`]

Therefore, `d/dx(x^n)=nx^{n-1}`.


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

Problem 2:   `d/dx(e^x)=e^x`

Solution:

Let `y=f(x)=e^x`

From the definition of derivatives, we get that 

`dy/dx=lim_{h to 0} frac{f(x+h)-f(x)}{h}` `=lim_{h to 0} frac{e^{x+h}-e^x}{h}`

`=lim_{h to 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` [`because lim_{x to 0}frac{e^x-1}{x}=1`]

`=e^x`

Therefore, `d/dx(e^x)=e^x`.

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

Problem 3:   `d/dx(a^x)=a^x log_e a`

Solution:

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

From the definition of derivatives, we get that 

`dy/dx=lim_{h to 0} frac{f(x+h)-f(x)}{h}` `=lim_{h to 0} frac{a^{x+h}-a^x}{h}`

`=lim_{h to 0} frac{a^x cdot a^h – a^x}{h}` `=a^x cdot lim_{h to 0} frac{a^h-1}{h}`

`=a^x cdot log_e a` [`because lim_{x to 0}frac{a^x-1}{x}=log_e a`]

`=a^x log_e a quad (a>0)`.

Therefore, `d/dx(e^x)=a^x log_e a`.

Next, we calculate the derivative of  `log x` by the definition of the derivative.

Problem 4:   `d/dx(log x)=1/x`

Solution:

Let `y=f(x)=log x`

From the definition of derivatives, we get that 

`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} 1/h log(frac{x+h}{x})` [`because log a – log b=log a/b`]

`=lim_{h to 0} 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, `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:   `d/dx(sin x)=cos x`

Solution:

Let `y=f(x)=sin x`

From the definition of derivatives, we get that 

`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=2cos frac{a+b}{2}sin frac{a-b}{2}`]

`=lim_{h to 0} 1/h cdot 2 cos frac{2x+h}{2}sin frac{h}{2}`  

`=lim_{h to 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 to 0} cos(x+h/2) cdot lim_{z to 0} frac{sin z}{z}`

`=cos(x+0) cdot 1 quad`[`because lim_{x to 0} sin x/x=1`]

`=cos x`

Therefore, `d/dx(sin x)=cos x`.


Next, we calculate the derivative of cos x by the definition of the derivative.

Problem 6:   `d/dx(cos x)=-sin x`

Solution:

Let `y=f(x)=cos x`

From the definition of derivatives, we get that 

`dy/dx=lim_{h to 0} frac{f(x+h)-f(x)}{h}` `=lim_{h to 0} frac{cos(x+h)-cos x}{h}`

[Formula used: ` cos a -cos b=2sin frac{a+b}{2}sin frac{b-a}{2}`]

`=lim_{h to 0} 1/h cdot 2 sin frac{2x+h}{2}sin frac{-h}{2}`  

`=-lim_{h to 0} sin(x+h/2) cdot lim_{h to 0} frac{sin h/2}{h/2}` [`because sin(-x)=-sin x`]

[Put `z=h/2`. Then `z to 0` as `h to 0`]

`=-lim_{h to 0} sin(x+h/2) cdot lim_{z to 0} frac{sin z}{z}`

`=-sin(x+0) cdot 1 quad`[`because lim_{x to 0} sin x/x=1`]

`=-sin x`

Therefore, `d/dx(cos x)=-sin x`.

Next, we calculate the derivative of tan x by the definition of the derivative.

Problem 7:   `d/dx(tan x)=sec^2 x`

Solution:

Let `y=f(x)=tan x`

From the definition of derivatives, we get that 

`dy/dx=lim_{h to 0} frac{f(x+h)-f(x)}{h}` `=lim_{h to 0} frac{tan(x+h)-tan x}{h}`

`=lim_{h to 0} 1/h [frac{sin(x+h)}{cos(x+h)}-frac{sin x}{cos x}]`  

`=lim_{h to 0}` `1/h [frac{sin(x+h)cos x – cos(x+h) sin x}{cos(x+h)cos x}]` 

`=lim_{h to 0} 1/h [frac{sin(x+h-x)}{cos(x+h) cos x}]`

`=lim_{h to 0} frac{sin h}{h} cdot lim_{h to 0} frac{1}{cos(x+h) cos x}`

[Formula used: `sin a cos b -cos a sin b = sin(a-b)`]

`=1 cdot frac{1}{cos x cosx} quad`[`because lim_{x to 0} sin x/x=1`]

`=sec^2 x` [`because sec x=frac{1}{cos x}`]

Therefore, `d/dx(cos x)=sec^2 x`.

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