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Assignment on Differentiation, find the derivation of the following with x using first principle: 1. F(x) = x2 − x 2. F(x) = sqrt(x+1) 3. g(x) = 1/x2 4. h(x) = 1/sqrt(x) 5. y = 1/(2x-1)

Mathematics

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Here are the derivations using the first principle:

  1. For F(x)=x2−xF(x) = x^2 - x: The first principle definition is F′(x)=lim⁡h→0F(x+h)−F(x)hF'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}.

  1. Find F(x+h)F(x+h).

    $$
F(x+h) = (x+h)^2 - (x+h) = x^2 + 2xh + h^2 - x - h
$$

  2. Find F(x+h)−F(x)F(x+h) - F(x).

    $$
F(x+h) - F(x) = (x^2 + 2xh + h^2 - x - h) - (x^2 - x) = 2xh + h^2 - h
$$

  3. Divide by hh.

    $$
\frac{F(x+h) - F(x)}{h} = \frac{2xh + h^2 - h}{h} = 2x + h - 1
$$

  4. Take the limit as h→0h \to 0.

    $$
F'(x) = \lim_{h \to 0} (2x + h - 1) = 2x - 1
$$
The derivative is $\boxed{2x-1}$.

    2. For F(x)=x+1F(x) = \sqrt{x+1}: The first principle definition is F′(x)=lim⁡h→0F(x+h)−F(x)hF'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}.

  5. Find F(x+h)−F(x)F(x+h) - F(x).

    $$
F(x+h) - F(x) = \sqrt{(x+h)+1} - \sqrt{x+1}
$$

  6. Multiply the numerator and denominator by the conjugate of the numerator.

    $$
\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \times \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}}
$$

  7. Simplify the numerator using (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2-b^2.

    $$
\frac{(x+h+1) - (x+1)}{h(\sqrt{x+h+1} + \sqrt{x+1})} = \frac{h}{h(\sqrt{x+h+1} + \sqrt{x+1})}
$$

  8. Cancel hh from the numerator and denominator.

    $$
\frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}
$$

  9. Take the limit as h→0h \to 0.

    $$
F'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} = \frac{1}{\sqrt{x+0+1} + \sqrt{x+1}} = \frac{1}{2\sqrt{x+1}}
$$
The derivative is $\boxed{\frac{1}{2\sqrt{x+1}}}$.

    3. For g(x)=1x2g(x) = \frac{1}{x^2}: The first principle definition is g′(x)=lim⁡h→0g(x+h)−g(x)hg'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}.

  10. Find g(x+h)−g(x)g(x+h) - g(x).

    $$
g(x+h) - g(x) = \frac{1}{(x+h)^2} - \frac{1}{x^2}
$$

  11. Combine the fractions in the numerator.

    $$
\frac{x^2 - (x+h)^2}{x^2(x+h)^2} = \frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2} = \frac{-2xh - h^2}{x^2(x+h)^2}
$$

  12. Divide by hh.

    $$
\frac{g(x+h) - g(x)}{h} = \frac{-2xh - h^2}{h \cdot x^2(x+h)^2} = \frac{h(-2x - h)}{h \cdot x^2(x+h)^2} = \frac{-2x - h}{x^2(x+h)^2}
$$

  13. Take the limit as h→0h \to 0.

    $$
g'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2} = \frac{-2x - 0}{x^2(x+0)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}
$$
The derivative is $\boxed{-\frac{2}{x^3}}$.

    4. For h(x)=1xh(x) = \frac{1}{\sqrt{x}}: The first principle definition is h′(x)=lim⁡h→0h(x+h)−h(x)hh'(x) = \lim_{h \to 0} \frac{h(x+h) - h(x)}{h}.

  14. Find h(x+h)−h(x)h(x+h) - h(x).

    $$
h(x+h) - h(x) = \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}
$$

  15. Combine the fractions in the numerator.

    $$
\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}
$$

  16. Multiply the numerator and denominator by the conjugate of the numerator.

    $$
\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}} = \frac{x - (x+h)}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \frac{-h}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
$$

  17. Divide by hh.

    $$
\frac{h(x+h) - h(x)}{h} = \frac{-h}{h \cdot \sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}
$$

  18. Take the limit as h→0h \to 0.

    $$
h'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \frac{-1}{\sqrt{x}\sqrt{x+0}(\sqrt{x} + \sqrt{x+0})} = \frac{-1}{\sqrt{x}\sqrt{x}(2\sqrt{x})} = \frac{-1}{x(2\sqrt{x})} = -\frac{1}{2x^{3/2}}
$$
The derivative is $\boxed{-\frac{1}{2x^{3/2}}}$.

    5. For y=12x−1y = \frac{1}{2x-1}: Let f(x)=12x−1f(x) = \frac{1}{2x-1}. The first principle definition is f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

  19. Find f(x+h)−f(x)f(x+h) - f(x).

    $$
f(x+h) - f(x) = \frac{1}{2(x+h)-1} - \frac{1}{2x-1} = \frac{1}{2x+2h-1} - \frac{1}{2x-1}
$$

  20. Combine the fractions in the numerator.

    $$
\frac{(2x-1) - (2x+2h-1)}{(2x+2h-1)(2x-1)} = \frac{2x-1-2x-2h+1}{(2x+2h-1)(2x-1)} = \frac{-2h}{(2x+2h-1)(2x-1)}
$$

  21. Divide by hh.

    $$
\frac{f(x+h) - f(x)}{h} = \frac{-2h}{h(2x+2h-1)(2x-1)} = \frac{-2}{(2x+2h-1)(2x-1)}
$$

  22. Take the limit as h→0h \to 0.

    $$
f'(x) = \lim_{h \to 0} \frac{-2}{(2x+2h-1)(2x-1)} = \frac{-2}{(2x+2(0)-1)(2x-1)} = \frac{-2}{(2x-1)(2x-1)} = \frac{-2}{(2x-1)^2}
$$
The derivative is $\boxed{-\frac{2}{(2x-1)^2}}$.
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