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`.