Question

Using first principle method, find the derivative of x^6

Answer 1

The derivative of a function f(x) is defined as the limit,

`f'(x):=\displaystyle\lim_(h\to0)\frac{f(x+h)-f(x)}h`

With f(x) = x⁶, we have

`f'(x)=\displaystyle\lim_(h\to0)\frac{(x+h)^6-x^6}h`

Expand f(x + h) in the numerator:

(x + h)⁶ = x⁶ + 6x⁵h + 15x⁴h² + 20x³h³ + 15x²h⁴ + 6xh⁵ + h⁶

so that the x⁶ terms cancel, leaving us with

`f'(x)=\displaystyle\lim_(h\to0)\frac{6x^5h+15x^4h^2+20x^3h^3+15x^2h^4+6xh^5+h^6}h`

h is approaching 0, so h ≠ 0 and we can cancel the common factor in the numerator and denominator:

`f'(x)=\displaystyle\lim_(h\to0)\left(6x^5+15x^4h+20x^3h^2+15x^2h^3+6xh^4+h^5\right)`

Now as h converges to 0, each term containing h vanishes, leaving us with

f'(x) = 6x⁵