Math Problem Statement

Find f'(x) by first principles for the functions: f(x) = 1/(x - 3) and f(x) = 1/(x + 2).

Solution

It looks like you're working on a calculus worksheet involving derivatives by first principles. I'll guide you through solving these problems.

First principles formula:

The derivative f(x)f'(x) by first principles is given by the limit:

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

Now let's approach both parts:

Part (a)

The function provided is:

f(x)=1x3f(x) = \frac{1}{x - 3}

Using the first principles, we need to evaluate:

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

Substitute f(x)f(x) and f(x+h)f(x+h):

f(x)=limh01(x+h)31x3hf'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)-3} - \frac{1}{x-3}}{h}

Simplify the numerator:

=limh01x+h31x3h=limh0(x3)(x+h3)(x+h3)(x3)1h= \lim_{h \to 0} \frac{\frac{1}{x+h-3} - \frac{1}{x-3}}{h} = \lim_{h \to 0} \frac{(x-3)-(x+h-3)}{(x+h-3)(x-3)} \cdot \frac{1}{h}

=limh0h(x+h3)(x3)h= \lim_{h \to 0} \frac{-h}{(x+h-3)(x-3) \cdot h}

Cancel the hh in the numerator and denominator:

=limh01(x+h3)(x3)= \lim_{h \to 0} \frac{-1}{(x+h-3)(x-3)}

As h0h \to 0, the expression simplifies to:

f(x)=1(x3)2f'(x) = \frac{-1}{(x-3)^2}

Part (b)

The function provided is:

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

Again, using the first principles:

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

Substitute f(x)f(x) and f(x+h)f(x+h):

f(x)=limh01x+h+21x+2hf'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h+2} - \frac{1}{x+2}}{h}

Simplify the numerator:

=limh0(x+2)(x+h+2)(x+h+2)(x+2)1h= \lim_{h \to 0} \frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)} \cdot \frac{1}{h}

=limh0h(x+h+2)(x+2)h= \lim_{h \to 0} \frac{-h}{(x+h+2)(x+2) \cdot h}

Cancel the hh:

=limh01(x+h+2)(x+2)= \lim_{h \to 0} \frac{-1}{(x+h+2)(x+2)}

As h0h \to 0, this simplifies to:

f(x)=1(x+2)2f'(x) = \frac{-1}{(x+2)^2}

Final answers:

  • (a) f(x)=1(x3)2f'(x) = \frac{-1}{(x-3)^2}
  • (b) f(x)=1(x+2)2f'(x) = \frac{-1}{(x+2)^2}

Would you like more detailed steps for either part, or explanations on a specific step?

Here are 5 related questions for practice:

  1. How does the first principles method differ from using derivative rules?
  2. Can you find the derivative of f(x)=1x+1f(x) = \frac{1}{x+1} using first principles?
  3. What is the derivative of f(x)=1x2+1f(x) = \frac{1}{x^2 + 1} by first principles?
  4. How would you approach finding the derivative of a quadratic function by first principles?
  5. Can you explain the concept of limits as used in the first principles definition of the derivative?

Tip: When working with limits in first principles, always aim to simplify the algebra before taking the limit to zero for cleaner results.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Limits
First Principles

Formulas

f'(x) = lim_{h -> 0} (f(x+h) - f(x)) / h
Derivative of f(x) = 1/(x - a) is -1/(x - a)^2

Theorems

First Principles of Derivatives

Suitable Grade Level

Grade 12 or introductory college calculus