To differentiate these functions, we'll use the definition of the derivative and then verify the results using shortcut rules (power, chain, and quotient rules). The definition of the derivative is:

$f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

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(a) $y = 3x - \sqrt{x}$

Using the definition:

1. Compute $f(x+\Delta x)$:
   $f(x+\Delta x) = 3(x+\Delta x) - \sqrt{x+\Delta x}$
2. Compute $f(x+\Delta x) - f(x)$:
   [ \left[3(x+\Delta x) - \sqrt{x+\Delta x} \right] - \left[3x - \sqrt{x} \right] ] $= 3x + 3\Delta x - \sqrt{x+\Delta x} - 3x + \sqrt{x}$ $= 3\Delta x - (\sqrt{x+\Delta x} - \sqrt{x})$
3. Divide by $\Delta x$ and take the limit:
   $f'(x) = \lim_{\Delta x \to 0} \frac{3\Delta x - (\sqrt{x+\Delta x} - \sqrt{x})}{\Delta x}$ Using the identity:
   $\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$ $\sqrt{x+\Delta x} - \sqrt{x} = \frac{(x+\Delta x) - x}{\sqrt{x+\Delta x} + \sqrt{x}} = \frac{\Delta x}{\sqrt{x+\Delta x} + \sqrt{x}}$ Thus,
   $f'(x) = \lim_{\Delta x \to 0} \left[ 3 - \frac{1}{\sqrt{x+\Delta x} + \sqrt{x}} \right]$ As $\Delta x \to 0$, $\sqrt{x+\Delta x} \to \sqrt{x}$, so: $f'(x) = 3 - \frac{1}{2\sqrt{x}}$

Using shortcuts:

• $\frac{d}{dx} (3x) = 3$
• $\frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}$

Thus,
$f'(x) = 3 - \frac{1}{2\sqrt{x}}$ ✔ Matches!

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(b) $y = (x^2 - 2)^2$

Using the definition (Omitted due to length; involves binomial expansion)

Using the chain rule

Let $u = x^2 - 2$, so $y = u^2$.

• $\frac{dy}{du} = 2u = 2(x^2 - 2)$
• $\frac{du}{dx} = 2x$
  By the chain rule:
  $f'(x) = 2(x^2 - 2) \cdot 2x = 4x(x^2 - 2)$

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(c) $y = 2 - 3x - \frac{1}{x}$

Using the definition (Omitted due to length)

Using shortcuts:

• $\frac{d}{dx} (2) = 0$
• $\frac{d}{dx} (-3x) = -3$
• $\frac{d}{dx} (-x^{-1}) = x^{-2} = \frac{1}{x^2}$

Thus,
$f'(x) = -3 + \frac{1}{x^2}$

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(d) $y = \frac{1}{\sqrt{x}} - 2$

Using the definition (Omitted due to length)

Using shortcuts:

Rewrite as $y = x^{-1/2} - 2$

• $\frac{d}{dx} (x^{-1/2}) = -\frac{1}{2} x^{-3/2} = -\frac{1}{2x^{3/2}}$
• $\frac{d}{dx} (-2) = 0$

Thus,
$f'(x) = -\frac{1}{2x^{3/2}}$

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Final Answers

1. $f'(x) = 3 - \frac{1}{2\sqrt{x}}$
2. $f'(x) = 4x(x^2 - 2)$
3. $f'(x) = -3 + \frac{1}{x^2}$
4. $f'(x) = -\frac{1}{2x^{3/2}}$

Do you want details on any step?

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Related Questions

1. How do you derive the power rule from the definition of the derivative?
2. How do you use the product rule in differentiation?
3. Can you explain the difference between the chain rule and the quotient rule?
4. What are some real-world applications of derivatives?
5. How do you find the second derivative using the chain rule?

Tip:

For square root functions, converting them to exponent form makes differentiation easier! 🚀