Derivative Of Root X Step-by-Step Solution

In the realm of calculus, derivatives play a pivotal role in understanding the rate at which a function changes. This article delves into the process of finding the derivative of the square root function, $f(x) = √x$, for all $x > 0$. This is a fundamental concept in calculus, with applications ranging from physics to economics. We will explore the formal definition of a derivative and apply it to this specific function, providing a clear and comprehensive guide for students and enthusiasts alike.

Understanding the Derivative

At its core, the derivative of a function at a given point represents the instantaneous rate of change of the function at that point. Geometrically, it corresponds to the slope of the tangent line to the function's graph at that point. The derivative is a cornerstone of calculus, enabling us to analyze the behavior of functions, including their increasing and decreasing intervals, concavity, and critical points.

To formally define the derivative, we employ the concept of a limit. The derivative of a function $f(x)$ at a point $x$ is denoted as $f'(x)$ and is defined as:

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

This limit, if it exists, gives us the instantaneous rate of change of $f(x)$ with respect to $x$. The expression $\frac{f(x + h) - f(x)}{h}$ represents the slope of a secant line passing through the points $(x, f(x))$ and $(x + h, f(x + h))$ on the graph of $f(x)$. As $h$ approaches 0, this secant line approaches the tangent line at the point $(x, f(x))$, and its slope approaches the derivative $f'(x)$.

The derivative is not merely an abstract mathematical concept; it has profound practical implications. In physics, it allows us to calculate velocity and acceleration from displacement functions. In economics, it helps us determine marginal cost and marginal revenue. In optimization problems, derivatives are crucial for finding maximum and minimum values of functions. Therefore, mastering the concept of derivatives is essential for anyone seeking a deeper understanding of mathematics and its applications.

Applying the Definition to f(x) = √x

Now, let's apply the definition of the derivative to find the derivative of $f(x) = √x$. This process involves substituting $f(x)$ into the limit definition and then manipulating the expression to evaluate the limit. Our goal is to find a function $f'(x)$ that represents the derivative of $√x$ at any point $x$ in its domain.

We start by plugging $f(x) = √x$ into the definition of the derivative:

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

At first glance, directly substituting $h = 0$ into the expression results in an indeterminate form $\frac{0}{0}$. To resolve this, we need to manipulate the expression algebraically. A common technique for dealing with square roots in limits is to multiply the numerator and denominator by the conjugate of the numerator. The conjugate of $\sqrt{x + h} - \sqrt{x}$ is $\sqrt{x + h} + \sqrt{x}$. Multiplying both the numerator and denominator by this conjugate, we get:

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

This step is crucial because it allows us to eliminate the square roots in the numerator. By multiplying by the conjugate, we are essentially using the difference of squares identity, which states that $(a - b)(a + b) = a^2 - b^2$. Applying this identity to the numerator, we obtain:

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

Simplifying the numerator, the $x$ terms cancel out, leaving us with:

$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}$

Now, we can cancel out the $h$ in the numerator and denominator, provided that $h ≠ 0$. This is a valid step because we are considering the limit as $h$ approaches 0, not the value of the expression at $h = 0$. This simplification yields:

$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$

We have successfully transformed the original limit expression into a form where we can now directly substitute $h = 0$ without encountering an indeterminate form. This is a significant milestone in our derivation.

Evaluating the Limit

With the simplified expression, we can now evaluate the limit by substituting $h = 0$:

$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{\sqrt{x + 0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}}$

Combining the terms in the denominator, we get:

$f'(x) = \frac{1}{2\sqrt{x}}$

Thus, we have found the derivative of $f(x) = √x$:

$f'(x) = \frac{1}{2\sqrt{x}}$

This result is valid for all $x > 0$, which is the domain of the original function $f(x) = √x$. The derivative, $f'(x) = \frac{1}{2\sqrt{x}}$, tells us the instantaneous rate of change of the square root function at any positive value of $x$. It also reveals that the slope of the tangent line to the graph of $y = √x$ decreases as $x$ increases, which aligns with the visual representation of the graph.

This derived formula is a cornerstone result in calculus and is frequently used in various applications. Understanding its derivation not only reinforces the concept of derivatives but also provides a valuable tool for solving more complex problems. The process of finding the derivative of $f(x) = √x$ exemplifies the power of the limit definition and algebraic manipulation in calculus.

Domain of the Derivative

An important consideration when working with derivatives is the domain over which the derivative is defined. Recall that the original function, $f(x) = √x$, is defined for all non-negative real numbers, i.e., $x ≥ 0$. However, when we found the derivative, we obtained the function:

$f'(x) = \frac{1}{2\sqrt{x}}$

This derivative is defined only for $x > 0$, not for $x = 0$. This is because the square root function in the denominator is not defined for negative numbers, and division by zero is undefined. Therefore, the domain of the derivative $f'(x)$ is $(0, ∞)$, which is a subset of the domain of the original function $f(x)$.

The fact that the derivative is not defined at $x = 0$ has a geometric interpretation. At $x = 0$, the graph of $f(x) = √x$ has a vertical tangent. Vertical lines have undefined slopes, which corresponds to the derivative being undefined at that point. This highlights an important principle: a function must be continuous and