Question

Find from first principles the first derivative of \((x+3)^{2}\) and compare your answer with that obtained using the chain rule.

Short answer

The derivative is \( 2x + 6 \). Both from first principles and the chain rule, the answer is \( 2x + 6 \).

Step-by-step solution

Understand the Derivative from First Principles

The first derivative from first principles is given by the limit definition:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]. Here, let \( f(x) = (x+3)^{2} \).

Substitute the Function into the Limit Definition

Substitute \( f(x) = (x+3)^2 \) into the limit definition formula:\[ f'(x) = \lim_{h \to 0} \frac{(x+h+3)^2 - (x+3)^2}{h} \].

Expand the Squares

Expand \((x+h+3)^2\) and \((x+3)^2\):\[ (x+h+3)^2 = x^2 + 6x + 9 + 2hx + h^2 + 6h \]\[ (x+3)^2 = x^2 + 6x + 9 \].

Simplify the Expression

Subtract the expanded forms and simplify the numerator:\[ (x^2 + 6x + 9 + 2hx + h^2 + 6h) - (x^2 + 6x + 9) \]\[ = 2hx + h^2 + 6h \].

Factor and Apply the Limit

Factor out \( h \) from the numerator:\[ f'(x) = \lim_{h \to 0} \frac{h(2x + h + 6)}{h} \]Cancel \( h \) and apply the limit:\[ f'(x) = \lim_{h \to 0} (2x + h + 6) \]\[ f'(x) = 2x + 6 \].

Verify Using the Chain Rule

To verify using the chain rule: let \( u = x + 3 \), then \( y = u^2 \) and \( \frac{dy}{du} = 2u \). Given \( u = x + 3 \), \( \frac{du}{dx} = 1 \). The chain rule gives:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot\frac{du}{dx} = 2u \cdot 1 = 2(x + 3) \]. Simplified, this is \( 2x + 6 \).

Key concepts

Limit Definition of Derivative

The concept of finding the derivative from first principles is based on the limit definition of the derivative. This is essentially the formal way to determine the rate at which a function changes at any given point. The formula for this is:\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]This formula tells us how the function behaves as the gap between the two points, defined by 'h', approaches zero. It gives us the slope of the tangent line to the function at the point x. To apply this formula, you need to:

• Substitute the function into the formula
• Simplify the expression inside the limit
• Evaluate the limit as 'h' approaches zero

Understanding this fundamental concept is crucial for differentiating functions from first principles.

Chain Rule

The chain rule is a powerful tool in calculus used for finding the derivative of composite functions. A composite function is created when one function is nested inside another, like \( f(g(x)) \). The chain rule formula is:\[ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \]

• Identify the inner function (u) and the outer function (y)
• Differentiate both functions separately
• Multiply the derivatives together

For example, to find the derivative of \((x+3)^2\), let \( u = x + 3 \). Then, \( y = u^2 \) and \( \frac{{dy}}{{du}} = 2u \). Since \( u = x + 3 \), \( \frac{{du}}{{dx}} = 1 \). So, applying the chain rule:\[ \frac{{dy}}{{dx}} = 2u \cdot 1 = 2(x + 3) \]This simplifies to \( 2x + 6 \), verifying our derivative obtained from first principles.

Expanding Algebraic Expressions

Expanding algebraic expressions is a fundamental step in many calculus problems. It involves rewriting an expression in an extended form by multiplying out the factors. For example, expanding \((x+3)^2\) involves:\[ (x+3)^2 = (x+3)(x+3) \to x^2 + 3x + 3x + 9 \to x^2 + 6x + 9 \]When substituting this into our limit definition, we also needed to expand \((x+h+3)^2\):\[ (x+h+3)^2 = (x+h+3)(x+h+3) = x^2 + 6x + 9 + 2hx + h^2 + 6h \]We then used these expanded forms to perform further simplification. Expanding expressions is key in transforming complex functions into simpler forms for easier manipulation.

Factorization

Factorization is the process of breaking down an algebraic expression into its simplest components, called factors. This is useful in simplifying expressions, especially when working with limits. In our derivative problem, after expanding and simplifying, we arrive at:\[ 2hx + h^2 + 6h \]To simplify further, we factor out 'h':\[ h(2x + h + 6) \]This allows us to cancel 'h' in the numerator and denominator:\[ \lim_{h \to 0} \frac{h(2x + h + 6)}{h} = \lim_{h \to 0} (2x + h + 6) \to 2x + 6 \]Factorization helps by simplifying complex terms, making the evaluation of limits and other operations straightforward.