FINDING FIRST DERIVATIVE USING FIRST PRINCIPLE

To find the slope function f'(x) for a general function f(x), we need to evaluate the limit.

h ⇾ 0 f(x + h) - f(x)h

We call this the method of first principle.

Find, from first principles, the derivative of f(x) equal to :

Problem 1 :

x + 2

Solution :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 x + h + 2 - x + 2hf'(x) = h ⇾ 0 x + h + 2 - x + 2h · x + h + 2 + x + 2x + h + 2 + x + 2= h ⇾ 0 x + h + 22 - x + 22hx + h + 2 + x + 2= h ⇾ 0 x + h + 2 - (x + 2)hx + h + 2 + x + 2= h ⇾ 0 x + h + 2 - x - 2hx + h + 2 + x + 2= h ⇾ 0 h hx + h + 2 + x + 2= h ⇾ 0 1 x + h + 2 + x + 2= 1x + 0 + 2 + x + 2= 1x + 2 + x + 2= 12x + 2

Problem 2 :

1x

Solution :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 1x + h - 1xhBy using cross multiplication.f'(x) = h ⇾ 0 x - x + hx · x + h hf'(x) = h ⇾ 0 x - x + hx · x + h 1hf'(x) = h ⇾ 0 x - x + hhx · x + h f'(x) = h ⇾ 0 x - x + hhx · x + h x + x + hx + x + hf'(x) = h ⇾ 0 x +xx + h -x + h x - (x + h) hx · x + h x + x + hf'(x) = h ⇾ 0 x +xx + h -x + h x - x - h hx · x + h x + x + hf'(x) = h ⇾ 0 - h hx · x + h x + x + hf'(x) = h ⇾ 0 - 1 x · x + h x + x + hf'(x) = -1x x x + xf'(x) = -1x 2x

Problem 3 :

2x + 1

Solution :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 2(x + h) + 1 - 2x + 1hf'(x) = h ⇾ 0 2(x + h) + 1 - 2x + 1h · 2(x + h) + 1 + 2x + 12(x + h) + 1 + 2x + 1= h ⇾ 0 2(x + h) + 12 - 2x + 12h2(x + h) + 1 + 2x + 1= h ⇾ 0 2(x + h) + 1 - (2x + 1)h2(x + h) + 1 + 2x + 1= h ⇾ 0 2x + 2h + 1 - 2x - 1h2(x + h) + 1 + 2x + 1= h ⇾ 0 2h h2(x + h) + 1 + 2x + 1= h ⇾ 0 2 2x + 2h + 1 + 2x + 1= 22x + 0 + 1 + 2x + 1= 22x + 1 + 2x + 1= 222x + 1 = 12x + 1

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