Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ g(x)=\sqrt{3 x} $$

Short answer

The derivative is \( g'(x) = \frac{1}{2\sqrt{x}} \).

Step-by-step solution

Understand the Problem

We are tasked with finding the derivative of the function \( g(x) = \sqrt{3x} \) at a general point \( x \) using the definition of a derivative. This involves applying the limit definition given as \( f^{\prime}(x)=\lim_{h \rightarrow 0} [f(x+h)-f(x)] / h \).

Set Up the Limit

Plug \( g(x) = \sqrt{3x} \) into the derivative definition. Thus, we have:\[g'(x) = \lim _{h \rightarrow 0}\frac{\sqrt{3(x+h)}-\sqrt{3x}}{h}\]

Simplify Using a Conjugate

Multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{3(x+h)} + \sqrt{3x} \), to simplify the expression:\[g'(x) = \lim _{h \rightarrow 0}\frac{\left(\sqrt{3(x+h)}-\sqrt{3x}\right)\left(\sqrt{3(x+h)}+\sqrt{3x}\right)}{h\left(\sqrt{3(x+h)}+\sqrt{3x}\right)}\]This becomes:\[g'(x) = \lim _{h \rightarrow 0}\frac{3(x+h) - 3x}{h\left(\sqrt{3(x+h)}+\sqrt{3x}\right)}\]

Cancel and Simplify

Simplify the expression by cancelling out terms in the numerator:\[g'(x) = \lim _{h \rightarrow 0}\frac{3h}{h\left(\sqrt{3(x+h)}+\sqrt{3x}\right)}\]Cancel the \( h \) in the numerator and denominator:\[g'(x) = \lim _{h \rightarrow 0}\frac{3}{\sqrt{3(x+h)}+\sqrt{3x}}\]

Evaluate the Limit

Evaluate the limit by substituting \( h = 0 \):\[g'(x) = \frac{3}{\sqrt{3x} + \sqrt{3x}} = \frac{3}{2\sqrt{3x}}\]Thus:\[g'(x) = \frac{1}{2\sqrt{x}}\]

State the Final Derivative

The derivative of \( g(x) = \sqrt{3x} \) with respect to \( x \) is \( g'(x) = \frac{1}{2\sqrt{x}} \).

Key concepts

limit definition

The concept of taking a derivative often starts with understanding the limit definition. This method allows us to calculate the derivative, which represents the slope of the tangent line at any point on a curve. The formula used is:

• \[ f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \]

In simpler terms, what we're doing is calculating how the function changes as we make small adjustments (denoted by \( h \)) to \( x \). By finding the limit as \( h \) approaches 0, we're finding this rate of change at exactly the point \( x \). The limit definition is powerful as it gives an exact result without approximations. It also forms the foundation for differentiation rules and techniques for more complex functions.

square root function

Square root functions, like \( \sqrt{3x} \), are a common type of function you'll encounter in calculus. They involve taking the square root of another function, here indicated by \( 3x \). Generally, the graph of a basic square root function, \( f(x) = \sqrt{x} \), has a shape called a 'half-parabola' which starts at the origin and curves upwards and to the right. It’s part of the family of radical functions.

• Growing at a decreasing rate: The rate of change (or slope) gets smaller as \( x \) increases.
• Domain and range considerations: The domain is \( x \geq 0 \), and the range is \( y \geq 0 \) for the basic \( f(x) = \sqrt{x} \), due to how square roots of negative numbers aren't real numbers in this context.

Knowing these characteristics helps us when graphing or analyzing these functions, as well as when applying calculus techniques like finding the derivative.

differentiation rules

Differentiation rules are formulas that show us the derivative of different types of functions quickly and easily. Instead of always resorting to the limit definition, these rules save time and effort. Key rules to know include the power rule, product rule, quotient rule, and chain rule.

• Power Rule: For \( f(x) = x^n \), the derivative is \( f^{\prime}(x) = n\cdot x^{n-1} \).
• Chain Rule: If a function is composed of another function, like \( \sqrt{3x} = (3x)^{1/2} \), then it's differentiated using the chain rule.

When using these rules on \( g(x) = \sqrt{3x} \), first rewrite it as \( (3x)^{1/2} \) and differentiate using the chain rule. You'd apply the power rule, as the derivative of \( x^{1/2} \) is \( \frac{1}{2} x^{-1/2} \), then multiply by the derivative of the inner function \( 3x \), which is 3. This yields our final derivative, \( g^{\prime}(x) = \frac{1}{2\sqrt{x}} \) after simplifying.