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The problem asks to find the derivative, $f'(x)$, of eight different functions, $f(x)$, using the definition of the derivative from first principles. The eight functions are: a) $f(x) = x - \frac{1}{x}$ b) $f(x) = \frac{1-x}{2+x}$ c) $f(x) = \frac{1}{x^2 - 3}$ d) $f(x) = |x|$ e) $f(x) = x^3 - x$ f) $f(x) = \sqrt{x}$ g) $f(x) = \sqrt{1 + 2x}$ h) $f(x) = x^3 + 2x$

AnalysisCalculusDifferentiationDerivativesLimitsFirst Principles
2025/5/12

1. Problem Description

The problem asks to find the derivative, f′(x)f'(x), of eight different functions, f(x)f(x), using the definition of the derivative from first principles. The eight functions are:
a) f(x)=x−1xf(x) = x - \frac{1}{x}
b) f(x)=1−x2+xf(x) = \frac{1-x}{2+x}
c) f(x)=1x2−3f(x) = \frac{1}{x^2 - 3}
d) f(x)=∣x∣f(x) = |x|
e) f(x)=x3−xf(x) = x^3 - x
f) f(x)=xf(x) = \sqrt{x}
g) f(x)=1+2xf(x) = \sqrt{1 + 2x}
h) f(x)=x3+2xf(x) = x^3 + 2x

2. Solution Steps

The definition of the derivative from first principles is given by:
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
a) f(x)=x−1xf(x) = x - \frac{1}{x}
f(x+h)=(x+h)−1x+hf(x+h) = (x+h) - \frac{1}{x+h}
f(x+h)−f(x)=(x+h)−1x+h−(x−1x)=h−1x+h+1x=h+x+h−xx(x+h)=h+hx(x+h)f(x+h) - f(x) = (x+h) - \frac{1}{x+h} - (x - \frac{1}{x}) = h - \frac{1}{x+h} + \frac{1}{x} = h + \frac{x+h-x}{x(x+h)} = h + \frac{h}{x(x+h)}
f(x+h)−f(x)h=1+1x(x+h)\frac{f(x+h) - f(x)}{h} = 1 + \frac{1}{x(x+h)}
f′(x)=lim⁡h→0(1+1x(x+h))=1+1x2f'(x) = \lim_{h \to 0} (1 + \frac{1}{x(x+h)}) = 1 + \frac{1}{x^2}
b) f(x)=1−x2+xf(x) = \frac{1-x}{2+x}
f(x+h)=1−(x+h)2+(x+h)f(x+h) = \frac{1-(x+h)}{2+(x+h)}
f(x+h)−f(x)=1−x−h2+x+h−1−x2+x=(1−x−h)(2+x)−(1−x)(2+x+h)(2+x+h)(2+x)=2+x−2x−x2−2h−xh−(2+x+h−2x−x2−xh)(2+x+h)(2+x)=−3h(2+x+h)(2+x)f(x+h) - f(x) = \frac{1-x-h}{2+x+h} - \frac{1-x}{2+x} = \frac{(1-x-h)(2+x) - (1-x)(2+x+h)}{(2+x+h)(2+x)} = \frac{2+x-2x-x^2-2h-xh - (2+x+h-2x-x^2-xh)}{(2+x+h)(2+x)} = \frac{-3h}{(2+x+h)(2+x)}
f(x+h)−f(x)h=−3(2+x+h)(2+x)\frac{f(x+h) - f(x)}{h} = \frac{-3}{(2+x+h)(2+x)}
f′(x)=lim⁡h→0−3(2+x+h)(2+x)=−3(2+x)2f'(x) = \lim_{h \to 0} \frac{-3}{(2+x+h)(2+x)} = \frac{-3}{(2+x)^2}
c) f(x)=1x2−3f(x) = \frac{1}{x^2 - 3}
f(x+h)=1(x+h)2−3f(x+h) = \frac{1}{(x+h)^2 - 3}
f(x+h)−f(x)=1(x+h)2−3−1x2−3=x2−3−((x+h)2−3)((x+h)2−3)(x2−3)=x2−3−(x2+2xh+h2−3)((x+h)2−3)(x2−3)=−2xh−h2((x+h)2−3)(x2−3)f(x+h) - f(x) = \frac{1}{(x+h)^2 - 3} - \frac{1}{x^2 - 3} = \frac{x^2 - 3 - ((x+h)^2 - 3)}{((x+h)^2 - 3)(x^2 - 3)} = \frac{x^2 - 3 - (x^2 + 2xh + h^2 - 3)}{((x+h)^2 - 3)(x^2 - 3)} = \frac{-2xh - h^2}{((x+h)^2 - 3)(x^2 - 3)}
f(x+h)−f(x)h=−2x−h((x+h)2−3)(x2−3)\frac{f(x+h) - f(x)}{h} = \frac{-2x - h}{((x+h)^2 - 3)(x^2 - 3)}
f′(x)=lim⁡h→0−2x−h((x+h)2−3)(x2−3)=−2x(x2−3)2f'(x) = \lim_{h \to 0} \frac{-2x - h}{((x+h)^2 - 3)(x^2 - 3)} = \frac{-2x}{(x^2 - 3)^2}
d) f(x)=∣x∣f(x) = |x|
This function is defined piecewise. We need to consider x>0x>0 and x<0x<0 separately.
If x>0x>0, f(x)=xf(x) = x, so f(x+h)=x+hf(x+h) = x+h. Then f′(x)=lim⁡h→0x+h−xh=lim⁡h→0hh=1f'(x) = \lim_{h \to 0} \frac{x+h - x}{h} = \lim_{h \to 0} \frac{h}{h} = 1.
If x<0x<0, f(x)=−xf(x) = -x, so f(x+h)=−(x+h)f(x+h) = -(x+h). Then f′(x)=lim⁡h→0−(x+h)−(−x)h=lim⁡h→0−hh=−1f'(x) = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} = \lim_{h \to 0} \frac{-h}{h} = -1.
At x=0x=0, the derivative does not exist.
Therefore f′(x)=1f'(x) = 1 if x>0x > 0 and f′(x)=−1f'(x) = -1 if x<0x < 0.
e) f(x)=x3−xf(x) = x^3 - x
f(x+h)=(x+h)3−(x+h)=x3+3x2h+3xh2+h3−x−hf(x+h) = (x+h)^3 - (x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - x - h
f(x+h)−f(x)=x3+3x2h+3xh2+h3−x−h−(x3−x)=3x2h+3xh2+h3−hf(x+h) - f(x) = x^3 + 3x^2h + 3xh^2 + h^3 - x - h - (x^3 - x) = 3x^2h + 3xh^2 + h^3 - h
f(x+h)−f(x)h=3x2+3xh+h2−1\frac{f(x+h) - f(x)}{h} = 3x^2 + 3xh + h^2 - 1
f′(x)=lim⁡h→0(3x2+3xh+h2−1)=3x2−1f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 - 1) = 3x^2 - 1
f) f(x)=xf(x) = \sqrt{x}
f(x+h)=x+hf(x+h) = \sqrt{x+h}
f(x+h)−f(x)=x+h−xf(x+h) - f(x) = \sqrt{x+h} - \sqrt{x}
Multiply by x+h+xx+h+x\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} to rationalize.
x+h−x=(x+h)−xx+h+x=hx+h+x\sqrt{x+h} - \sqrt{x} = \frac{(x+h) - x}{\sqrt{x+h} + \sqrt{x}} = \frac{h}{\sqrt{x+h} + \sqrt{x}}
f(x+h)−f(x)h=1x+h+x\frac{f(x+h) - f(x)}{h} = \frac{1}{\sqrt{x+h} + \sqrt{x}}
f′(x)=lim⁡h→01x+h+x=12xf'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}
g) f(x)=1+2xf(x) = \sqrt{1 + 2x}
f(x+h)=1+2(x+h)=1+2x+2hf(x+h) = \sqrt{1 + 2(x+h)} = \sqrt{1 + 2x + 2h}
f(x+h)−f(x)=1+2x+2h−1+2xf(x+h) - f(x) = \sqrt{1 + 2x + 2h} - \sqrt{1 + 2x}
Multiply by 1+2x+2h+1+2x1+2x+2h+1+2x\frac{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}} to rationalize.
1+2x+2h−1+2x=(1+2x+2h)−(1+2x)1+2x+2h+1+2x=2h1+2x+2h+1+2x\sqrt{1 + 2x + 2h} - \sqrt{1 + 2x} = \frac{(1 + 2x + 2h) - (1 + 2x)}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}} = \frac{2h}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}
f(x+h)−f(x)h=21+2x+2h+1+2x\frac{f(x+h) - f(x)}{h} = \frac{2}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}
f′(x)=lim⁡h→021+2x+2h+1+2x=221+2x=11+2xf'(x) = \lim_{h \to 0} \frac{2}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}} = \frac{2}{2\sqrt{1+2x}} = \frac{1}{\sqrt{1+2x}}
h) f(x)=x3+2xf(x) = x^3 + 2x
f(x+h)=(x+h)3+2(x+h)=x3+3x2h+3xh2+h3+2x+2hf(x+h) = (x+h)^3 + 2(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h
f(x+h)−f(x)=x3+3x2h+3xh2+h3+2x+2h−(x3+2x)=3x2h+3xh2+h3+2hf(x+h) - f(x) = x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h - (x^3 + 2x) = 3x^2h + 3xh^2 + h^3 + 2h
f(x+h)−f(x)h=3x2+3xh+h2+2\frac{f(x+h) - f(x)}{h} = 3x^2 + 3xh + h^2 + 2
f′(x)=lim⁡h→0(3x2+3xh+h2+2)=3x2+2f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 + 2) = 3x^2 + 2

3. Final Answer

a) f′(x)=1+1x2f'(x) = 1 + \frac{1}{x^2}
b) f′(x)=−3(2+x)2f'(x) = \frac{-3}{(2+x)^2}
c) f′(x)=−2x(x2−3)2f'(x) = \frac{-2x}{(x^2 - 3)^2}
d) f′(x)=1f'(x) = 1 if x>0x > 0, f′(x)=−1f'(x) = -1 if x<0x < 0. The derivative is undefined at x=0x=0.
e) f′(x)=3x2−1f'(x) = 3x^2 - 1
f) f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}
g) f′(x)=11+2xf'(x) = \frac{1}{\sqrt{1+2x}}
h) f′(x)=3x2+2f'(x) = 3x^2 + 2

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