FIND THE DERIVATIVES USING FIRST PRINCIPAL

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, f'(x) given that f(x) is :

Problem 1 :

2x + 5

Solution :

Let f(x) = 2x + 5

Derivative of (2x + 5) by first principle :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 2(x + h) + 5 - (2x + 5)hf'(x) = h ⇾ 0 2x + 2h + 5 - 2x - 5hh ⇾ 0 2hh= 2

(2x + 5)' = 2

Problem 2 :

x2 - 3x

Solution :

Let f(x) = x2 - 3x

Derivative of (x2 - 3x) by first principle :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 (x + h)2 - 3(x + h) - x2- 3xhf'(x) = h ⇾ 0 x2 + 2xh + h2 - 3x - 3h - x2 +3xhf'(x) = h ⇾ 0 2xh + h2 - 3hhf'(x) = h ⇾ 0 2x + h - 3

(x2 - 3x)' = 2x - 3

Problem 3 :

x3 - 2x2 + 3

Solution :

Let f(x) = x3 - 2x2 + 3

Derivative of (x3 - 2x2 + 3) by first principle :

f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 (x + h)3 - 2(x + h)2 + 3 -x3 - 2x2+ 3hf'(x) = h ⇾ 0 x3 + 3x2h + 3h2x + h3 - 2x2 + 2xh + h2 + 3 - x3 + 2x2 -3hf'(x) = h ⇾ 0 x3 + 3xh(x + h) + h3 - 2x2 - 4xh - 2h2+ 3 - x3 + 2x2 -3hf'(x) = h ⇾ 0 3xh(x + h) + h3 - 4xh - 2h2hf'(x) = h ⇾ 0 3x(x + h) + h2 - 4x - 2hf'(x) = h ⇾ 0 3x2 + 3xh + h2 - 4x - 2h

(x3 - 2x2 + 3)' = 3x2 - 4x

Find, from first principles, the derivative of f(x) where f(x) is :

Problem 4 :

1x + 2

Solution :

Let f(x) = 1x + 2Derivative of 1x + 2 by first principle :f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 1(x + h) + 2 - 1x + 2hBy using cross multiplication.f'(x) = h ⇾ 0 (x + 2) - ((x + h) + 2)(x + h + 2) · (x + 2)hf'(x) = h ⇾ 0 x + 2 - x - h - 2(x + h + 2) · (x + 2)hf'(x) = h ⇾ 0 - h (x + h + 2) · (x + 2)hf'(x) = h ⇾ 0 -h(x + h + 2) · (x + 2) · 1hf'(x) = h ⇾ 0 -1(x + h + 2) · (x + 2)= -1(x + 2) (x + 2)1x + 2'=-1(x + 2)2

Problem 5 :

12x - 1

Solution :

Let f(x) = 12x - 1Derivative of 12x - 1 by first principle :f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 12(x + h) - 1 - 12x - 1hBy using cross multiplication.f'(x) = h ⇾ 0 (2x - 1) - (2(x + h) - 1)2(x + h) - 1 · (2x - 1)hf'(x) = h ⇾ 0 2x - 1 - (2x + 2h - 1)2(x + h) - 1 · (2x - 1)hf'(x) = h ⇾ 0 2x - 1 - 2x - 2h + 1 2(x + h) - 1 · (2x - 1)hf'(x) = h ⇾ 0 -2h2(x + h) - 1 · (2x - 1) · 1hf'(x) = h ⇾ 0 -22(x + h) - 1 · (2x - 1)= -22(x + 0) - 1 ·(2x - 1)= -2(2x - 1) · (2x - 1)12x - 1'=-2(2x - 1)2

Problem 6 :

1x2

Solution :

Let f(x) = 1x2Derivative of 1x2 by first principle :f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 1(x + h)2 - 1x2hBy using cross multiplication.f'(x) = h ⇾ 0 x2 - (x + h)2(x + h)2 · x2 hf'(x) = h ⇾ 0 (x2 - x2 + h2 + 2xh(x + h)2 · x2 hf'(x) = h ⇾ 0 x2 - x2 - h2 - 2xh(x + h)2 · x2 hf'(x) = h ⇾ 0 - h2 - 2xh(x + h)2 · x2 hf'(x) = h ⇾ 0 -h2 - 2xh(x + h)2· x2 · 1hf'(x) = h ⇾ 0 h(-h - 2x)(x + h)2 · x2 · 1h f'(x) = h ⇾ 0 -h - 2x(x + h)2 · x2f'(x) = -2x(x + 0)2 · x2f'(x) = -2xx2 · x2f'(x) = -2xx4 1x2'= -2x3

Problem 7 :

1x3

Solution :

Let f(x) = 1x3Derivative of 1x3 by first principle :f'(x) = h ⇾ 0 f(x + h) - f(x)hf'(x) = h ⇾ 0 1(x + h)3 - 1x3hBy using cross multiplication.f'(x) = h ⇾ 0 x3 - (x + h)3(x + h)3 · x3 hf'(x) = h ⇾ 0 (x3 - x3 + h3 + 3x2h + 3h2x(x + h)3 · x3 hf'(x) = h ⇾ 0 x3 - x3 - h3 - 3x2h - 3h2x(x + h)3 · x3 hf'(x) = h ⇾ 0 - h3 - 3x2h - 3h2x(x + h)3 · x3 hf'(x) = h ⇾ 0 -h3 - 3x2h - 3h2x(x + h)3· x3 · 1hf'(x) = h ⇾ 0 h-h2 - 3x2 - 3hx(x + h)3 · x3 · 1h f'(x) = h ⇾ 0 -h2 - 3x2 - 3hx(x + h)3 · x3f'(x) = -3x2(x + 0)3 · x3f'(x) = -3x2x3 · x3f'(x) = -3x2x6 1x3'= -3x4

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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