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

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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 + 6xh + 15xh² + 20x³h³ + 15x²h⁴ + 6xh⁵ + h

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) = 6x

User Faryn
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