Math Problem Statement

If f(x) = √(2 + 2x), find f'(3) using the definition of derivative.

Solution

The problem asks to find f(3)f'(3) using the definition of the derivative, where the function is given as:

f(x)=2+2xf(x) = \sqrt{2 + 2x}

The definition of the derivative is:

f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

We need to calculate f(3)f'(3), so a=3a = 3. Let’s break this down step by step.

Step 1: Plug f(x)f(x) into the derivative formula

We know that f(x)=2+2xf(x) = \sqrt{2 + 2x}. For f(3)f'(3), we use:

f(3)=limh0f(3+h)f(3)hf'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}

Step 2: Calculate f(3)f(3)

f(3)=2+2(3)=2+6=8=22f(3) = \sqrt{2 + 2(3)} = \sqrt{2 + 6} = \sqrt{8} = 2\sqrt{2}

Step 3: Calculate f(3+h)f(3+h)

f(3+h)=2+2(3+h)=2+6+2h=8+2hf(3+h) = \sqrt{2 + 2(3+h)} = \sqrt{2 + 6 + 2h} = \sqrt{8 + 2h}

Step 4: Apply to the derivative formula

Substitute into the derivative formula:

f(3)=limh08+2h22hf'(3) = \lim_{h \to 0} \frac{\sqrt{8 + 2h} - 2\sqrt{2}}{h}

Step 5: Simplify the expression using the conjugate

To simplify, multiply the numerator and denominator by the conjugate of the numerator:

8+2h22h8+2h+228+2h+22=(8+2h)(8)h(8+2h+22)\frac{\sqrt{8 + 2h} - 2\sqrt{2}}{h} \cdot \frac{\sqrt{8 + 2h} + 2\sqrt{2}}{\sqrt{8 + 2h} + 2\sqrt{2}} = \frac{(8 + 2h) - (8)}{h(\sqrt{8 + 2h} + 2\sqrt{2})}

Simplifying the numerator:

2hh(8+2h+22)=28+2h+22\frac{2h}{h(\sqrt{8 + 2h} + 2\sqrt{2})} = \frac{2}{\sqrt{8 + 2h} + 2\sqrt{2}}

Step 6: Take the limit as h0h \to 0

As h0h \to 0, 8+2h8=22\sqrt{8 + 2h} \to \sqrt{8} = 2\sqrt{2}. Therefore:

f(3)=222+22=242=122=24f'(3) = \frac{2}{2\sqrt{2} + 2\sqrt{2}} = \frac{2}{4\sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}

Thus, the derivative at x=3x = 3 is:

f(3)=24f'(3) = \frac{\sqrt{2}}{4}

Would you like further clarification or details on any steps?

Here are five related questions for practice:

  1. How would you differentiate a function like f(x)=3+5xf(x) = \sqrt{3 + 5x} at x=2x = 2?
  2. Can you apply the quotient rule to find the derivative of a similar function?
  3. How does the conjugate method simplify the calculation in the derivative definition?
  4. What is the geometric interpretation of the derivative at a point like x=3x = 3?
  5. Can the derivative of a square root function be found using chain rule instead of the definition?

Tip: Always consider rationalizing expressions involving square roots when simplifying limits in derivatives.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Limits

Formulas

f'(a) = lim(h -> 0) [f(a+h) - f(a)] / h

Theorems

Definition of the Derivative

Suitable Grade Level

Undergraduate Calculus or AP Calculus