Derivative Of Y = Ln(1/(x√(x+9))) A Step-by-Step Solution
In the realm of calculus, derivatives serve as a fundamental tool for understanding the rate at which a function changes. They provide insights into the slope of a curve at any given point, enabling us to analyze the behavior of functions and model real-world phenomena. One common task in calculus is finding the derivative of a given function, which involves applying various differentiation rules and techniques. In this comprehensive guide, we will embark on a step-by-step journey to find the derivative of the function y = ln(1/(x√(x+9))) with respect to x. This exploration will not only enhance your understanding of differentiation but also showcase the power of calculus in unraveling the intricacies of mathematical expressions.
1. Simplifying the Function: Laying the Groundwork for Differentiation
Before we delve into the process of differentiation, it's often beneficial to simplify the function as much as possible. This simplification makes the subsequent differentiation steps more manageable and reduces the likelihood of errors. Our function, y = ln(1/(x√(x+9))), can be simplified using the properties of logarithms. The key property we'll employ here is the logarithm of a quotient, which states that ln(a/b) = ln(a) - ln(b). Applying this property to our function, we get:
y = ln(1) - ln(x√(x+9))
Since ln(1) = 0, our function simplifies to:
y = -ln(x√(x+9))
Next, we can utilize another property of logarithms, the logarithm of a product, which states that ln(ab) = ln(a) + ln(b). Applying this property, we have:
y = -[ln(x) + ln(√(x+9))]
Distributing the negative sign, we get:
y = -ln(x) - ln(√(x+9))
Finally, we can rewrite the square root as a power of 1/2, giving us:
y = -ln(x) - ln((x+9)^(1/2))
We can further simplify this expression using the power rule of logarithms, which states that ln(a^b) = b ln(a). Applying this rule, we obtain:
y = -ln(x) - (1/2)ln(x+9)
Now, our function is in a much simpler form, making it easier to differentiate. This initial simplification step is crucial as it streamlines the subsequent steps and minimizes the risk of making errors during differentiation. The simplified form allows us to directly apply the differentiation rules for logarithmic functions, which we will explore in the next section.
2. Applying Differentiation Rules: Unveiling the Derivative
With our function simplified to y = -ln(x) - (1/2)ln(x+9), we can now proceed to find its derivative with respect to x. To do this, we will employ the following differentiation rules:
- Derivative of ln(x): The derivative of ln(x) with respect to x is 1/x.
- Chain Rule: The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x).
- Constant Multiple Rule: The derivative of a constant multiplied by a function, cf(x), is cf'(x).
Applying these rules, we can differentiate each term of our function separately.
First, let's differentiate -ln(x). Using the derivative of ln(x) rule and the constant multiple rule, we get:
d/dx [-ln(x)] = -d/dx [ln(x)] = -1/x
Next, let's differentiate -(1/2)ln(x+9). This requires the chain rule. Let u = x+9. Then, our term becomes -(1/2)ln(u). Applying the chain rule, we have:
d/dx [-(1/2)ln(x+9)] = -(1/2) * d/du [ln(u)] * du/dx
The derivative of ln(u) with respect to u is 1/u, and the derivative of u = x+9 with respect to x is 1. Substituting these values, we get:
d/dx [-(1/2)ln(x+9)] = -(1/2) * (1/(x+9)) * 1 = -1/(2(x+9))
Now, we can combine the derivatives of the two terms to find the derivative of the entire function:
dy/dx = d/dx [-ln(x)] + d/dx [-(1/2)ln(x+9)]
dy/dx = -1/x - 1/(2(x+9))
This expression represents the derivative of our function y = -ln(x) - (1/2)ln(x+9) with respect to x. However, we can further simplify this expression to obtain a more compact form, which we will do in the next section.
3. Simplifying the Derivative: Expressing the Result in its Elegant Form
Our derivative, dy/dx = -1/x - 1/(2(x+9)), can be simplified by finding a common denominator and combining the fractions. The common denominator for the two terms is 2x(x+9). Multiplying the first term by 2(x+9)/2(x+9) and the second term by x/x, we get:
dy/dx = -[2(x+9)]/[2x(x+9)] - x/[2x(x+9)]
Combining the numerators, we have:
dy/dx = -[2(x+9) + x]/[2x(x+9)]
Expanding the numerator, we get:
dy/dx = -(2x + 18 + x)/[2x(x+9)]
Simplifying the numerator, we obtain:
dy/dx = -(3x + 18)/[2x(x+9)]
We can further simplify this expression by factoring out a 3 from the numerator:
dy/dx = -3(x + 6)/[2x(x+9)]
This is the simplified form of the derivative of our function. It expresses the rate of change of y with respect to x in a concise and elegant manner. This simplified form is not only aesthetically pleasing but also easier to work with in further calculations or analysis.
4. Applications of the Derivative: Unveiling the Power of Calculus
The derivative we found, dy/dx = -3(x + 6)/[2x(x+9)], is not just a mathematical expression; it holds valuable information about the behavior of the original function, y = ln(1/(x√(x+9))). Derivatives have a wide range of applications in various fields, including physics, engineering, economics, and computer science. Let's explore some of the key applications of our derivative.
4.1. Finding Critical Points: Identifying Maxima and Minima
Critical points of a function are points where the derivative is either zero or undefined. These points are crucial because they often correspond to local maxima, local minima, or saddle points of the function. To find the critical points of our function, we need to set the derivative equal to zero and solve for x:
-3(x + 6)/[2x(x+9)] = 0
A fraction is equal to zero only if its numerator is zero. Therefore, we have:
-3(x + 6) = 0
Solving for x, we get:
x = -6
Now, we need to check where the derivative is undefined. The derivative is undefined when the denominator is zero:
2x(x+9) = 0
This gives us two solutions:
x = 0 and x = -9
Thus, we have three potential critical points: x = -6, x = 0, and x = -9. However, we need to consider the domain of the original function, y = ln(1/(x√(x+9))). The function is defined only when the argument of the logarithm is positive, which means:
1/(x√(x+9)) > 0
This inequality holds when x > 0 or -9 < x < 0. Therefore, x = -6 and x = -9 are not in the domain of the function, and the only critical point we need to consider is x = 0. However, x = 0 is also not in the domain. This means that there are no critical points in the domain of the function.
4.2. Determining Intervals of Increase and Decrease: Understanding Function Behavior
The sign of the derivative tells us whether the function is increasing or decreasing. If the derivative is positive, the function is increasing, and if the derivative is negative, the function is decreasing. To determine the intervals of increase and decrease, we can analyze the sign of the derivative in different intervals of the domain.
Our domain is x > 0. Let's choose a test value in this interval, say x = 1. Plugging this value into the derivative, we get:
dy/dx = -3(1 + 6)/[2(1)(1+9)] = -21/20 < 0
Since the derivative is negative for x = 1, it is negative for all x > 0. Therefore, the function is decreasing in the interval (0, ∞).
4.3. Finding Concavity and Inflection Points: Unveiling the Curvature
The second derivative of a function provides information about its concavity. If the second derivative is positive, the function is concave up, and if the second derivative is negative, the function is concave down. Inflection points are points where the concavity changes.
To find the second derivative, we need to differentiate our first derivative, dy/dx = -3(x + 6)/[2x(x+9)]. This requires the quotient rule, which states that the derivative of u/v is (vu' - uv')/v^2**. Let u = -3(x+6) and v = 2x(x+9). Then, we have:
u' = -3 and v' = 2(2x + 9)
Applying the quotient rule, we get:
d2y/dx2 = [2x(x+9)(-3) - (-3(x+6))(2(2x+9))]/[2x(x+9)]^2
Simplifying this expression, we obtain:
d2y/dx2 = [6(x^2 + 12x + 27)]/[4x2(x+9)2]
d2y/dx2 = [3(x^2 + 12x + 27)]/[2x2(x+9)2]
Now, we need to analyze the sign of the second derivative. The denominator is always positive for x > 0. The numerator is a quadratic expression, x^2 + 12x + 27, which can be factored as (x+3)(x+9). Since x > 0, both factors are positive, and the numerator is positive. Therefore, the second derivative is positive for all x > 0, which means the function is concave up in the interval (0, ∞).
Since the concavity does not change, there are no inflection points.
4.4. Optimization Problems: Finding the Best Solution
Derivatives are essential tools for solving optimization problems, where the goal is to find the maximum or minimum value of a function. In many real-world scenarios, we need to optimize certain quantities, such as profit, cost, or area. Derivatives provide a systematic way to find these optimal values.
While our specific function may not have a practical optimization application, the principles we discussed in finding critical points and intervals of increase and decrease are fundamental to solving optimization problems. By setting the derivative equal to zero and analyzing the sign of the derivative, we can identify the points where the function reaches its maximum or minimum values.
Conclusion: A Journey Through Differentiation and its Applications
In this comprehensive guide, we have embarked on a journey to find the derivative of the function y = ln(1/(x√(x+9))). We started by simplifying the function using the properties of logarithms, making it easier to differentiate. Then, we applied the differentiation rules, including the chain rule and the derivative of ln(x), to find the derivative. We further simplified the derivative to obtain a concise expression.
Finally, we explored the applications of the derivative, including finding critical points, determining intervals of increase and decrease, finding concavity and inflection points, and solving optimization problems. These applications highlight the power of calculus in understanding the behavior of functions and modeling real-world phenomena.
This exploration has not only provided a step-by-step solution to finding the derivative of a specific function but also underscored the broader significance of derivatives in calculus and its applications. By mastering the techniques and concepts discussed in this guide, you will be well-equipped to tackle a wide range of differentiation problems and unlock the power of calculus in various fields of study and practice.
