Derivative Of Y = 10^x / (x + 2) Calculation Guide

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In the realm of calculus, derivatives stand as a cornerstone, offering profound insights into the rates at which functions change. This article delves into the intricacies of finding the derivative of a specific function, $y=\frac{10^x}{x+2}$, a task that blends exponential and rational expressions. We will embark on a step-by-step journey, unraveling the application of the quotient rule and chain rule, ultimately revealing the derivative, $\frac{dy}{dx}$. This exploration not only provides a solution but also aims to enhance your understanding of differentiation techniques, making you more adept at tackling similar calculus challenges.

Decoding the Derivative: A Journey Through Calculus

To find the derivative of $y=\frac{10^x}{x+2}$, we must employ a combination of calculus techniques, primarily the quotient rule and the chain rule. The quotient rule is indispensable when dealing with functions expressed as a ratio, while the chain rule comes into play when differentiating composite functions. Before we dive into the mathematical manipulations, let's lay a solid foundation by revisiting these essential rules.

The quotient rule states that the derivative of a quotient of two functions, $u(x)$ and $v(x)$, is given by:

ddx(u(x)v(x))=v(x)u′(x)−u(x)v′(x)[v(x)]2\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}

In simpler terms, it dictates that the derivative of a fraction is found by taking the denominator times the derivative of the numerator, subtracting the numerator times the derivative of the denominator, and then dividing the entire expression by the square of the denominator. This rule is a fundamental tool in calculus, allowing us to differentiate a wide range of rational functions.

Next, we encounter the chain rule, which is crucial for differentiating composite functions. A composite function is essentially a function within a function, such as $f(g(x))$. The chain rule provides a way to find the derivative of such functions, stating that:

ddxf(g(x))=f′(g(x))⋅g′(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

In essence, the chain rule tells us to differentiate the outer function, keeping the inner function unchanged, and then multiply by the derivative of the inner function. This rule is particularly important when dealing with exponential functions, trigonometric functions, and other complex expressions.

With these rules firmly in mind, we can now approach the differentiation of $y=\frac{10^x}{x+2}$. We'll identify the numerator and denominator, apply the quotient rule, and then use the chain rule where necessary. This process will not only yield the correct answer but also reinforce your understanding of these fundamental calculus concepts.

Applying the Quotient Rule: A Step-by-Step Differentiation

Now, let's apply the quotient rule to our function, $y=\frac{10^x}{x+2}$. We'll identify $u(x) = 10^x$ as the numerator and $v(x) = x+2$ as the denominator. According to the quotient rule, we need to find the derivatives of both $u(x)$ and $v(x)$.

Let's start with the derivative of $u(x) = 10^x$. This is where the chain rule subtly enters the picture. We know that the derivative of $a^x$ is $a^x \ln(a)$, where $a$ is a constant. In our case, $a = 10$, so the derivative of $10^x$ is $10^x \ln(10)$. This result stems from the chain rule applied to the exponential function, where the outer function is the exponential and the inner function is simply $x$.

Moving on to the derivative of $v(x) = x+2$, this is a straightforward application of the power rule and the constant rule. The derivative of $x$ is 1, and the derivative of the constant 2 is 0. Therefore, the derivative of $x+2$ is simply 1. This step might seem trivial, but it's crucial to ensure accuracy in the overall calculation.

Now that we have the derivatives of both the numerator and the denominator, we can plug them into the quotient rule formula:

dydx=(x+2)(10xln⁡(10))−(10x)(1)(x+2)2\frac{dy}{dx} = \frac{(x+2)(10^x \ln(10)) - (10^x)(1)}{(x+2)^2}

This expression represents the derivative of our original function. However, we can simplify it further to make it more elegant and easier to work with. Simplification often involves factoring and combining like terms, a crucial skill in calculus and other areas of mathematics. In the next section, we'll focus on simplifying this expression to its most concise form.

Simplifying the Derivative: Unveiling the Final Result

Having applied the quotient rule, we've arrived at the expression:

dydx=(x+2)(10xln⁡(10))−(10x)(1)(x+2)2\frac{dy}{dx} = \frac{(x+2)(10^x \ln(10)) - (10^x)(1)}{(x+2)^2}

Our next step is to simplify this expression. Simplification not only makes the result more aesthetically pleasing but also often makes it easier to use in subsequent calculations or analyses. The key to simplifying this expression lies in recognizing common factors and strategically factoring them out.

Observe that $10^x$ appears in both terms in the numerator. This suggests that we can factor out $10^x$ from the numerator. Doing so, we get:

dydx=10x[(x+2)ln⁡(10)−1](x+2)2\frac{dy}{dx} = \frac{10^x[(x+2)\ln(10) - 1]}{(x+2)^2}

This factored form is significantly cleaner than the previous expression. It clearly isolates the exponential term and groups the remaining terms within the brackets. This form is not only more concise but also provides insights into the behavior of the derivative. For instance, we can see that the derivative is directly proportional to $10^x$, which means it will always be positive and will grow exponentially as $x$ increases.

The expression inside the brackets, $(x+2)\ln(10) - 1$, cannot be simplified further without resorting to approximations or numerical methods. However, it provides valuable information about the critical points of the function, where the derivative is zero or undefined. Setting this expression equal to zero and solving for $x$ would give us the points where the function has a horizontal tangent, which could correspond to local maxima or minima.

Therefore, the simplified derivative of $y=\frac{10^x}{x+2}$ is:

dydx=10x[(x+2)ln⁡(10)−1](x+2)2\frac{dy}{dx} = \frac{10^x[(x+2)\ln(10) - 1]}{(x+2)^2}

This is our final result, a compact and informative expression that captures the rate of change of the original function. This result encapsulates the application of the quotient rule, the chain rule, and algebraic simplification techniques, showcasing the power and elegance of calculus.

Conclusion: Mastering Differentiation Techniques

In this article, we embarked on a journey to find the derivative of the function $y=\frac{10^x}{x+2}$. This seemingly simple task required the application of fundamental calculus concepts, including the quotient rule and the chain rule. We meticulously dissected the problem, identified the relevant rules, and executed the differentiation process step by step.

We began by laying the groundwork, understanding the quotient rule and the chain rule in their full glory. These rules are not merely formulas to be memorized; they are powerful tools that allow us to tackle a wide variety of differentiation problems. The quotient rule handles functions expressed as ratios, while the chain rule addresses composite functions, where one function is nested within another.

Next, we applied the quotient rule to our specific function, carefully identifying the numerator and the denominator. This involved differentiating both parts, which in turn required the application of the chain rule to the exponential term. The derivative of $10^x$ is a classic example of the chain rule in action, highlighting the importance of recognizing composite functions.

Once we obtained the initial derivative, we focused on simplification. Simplification is a crucial step in calculus, as it often leads to more manageable and insightful expressions. We factored out common terms, revealing the underlying structure of the derivative and making it easier to interpret. The simplified derivative, $\frac{dy}{dx} = \frac{10^x[(x+2)\ln(10) - 1]}{(x+2)^2}$, is a testament to the power of algebraic manipulation in calculus.

This exercise not only provides the answer to a specific differentiation problem but also serves as a valuable learning experience. It reinforces the importance of understanding the underlying principles of calculus, mastering the application of differentiation rules, and honing algebraic simplification skills. By tackling this problem, you've strengthened your ability to approach similar calculus challenges with confidence and competence.

The world of calculus is vast and fascinating, filled with intricate concepts and powerful tools. Mastering these concepts and tools opens doors to understanding a wide range of phenomena, from the motion of objects to the growth of populations. Differentiation, in particular, is a cornerstone of calculus, providing insights into rates of change and laying the foundation for more advanced topics such as integration and differential equations. So, continue exploring, continue practicing, and continue unraveling the mysteries of calculus.

This article provides a step-by-step guide on how to calculate the derivative of the function y = 10^x / (x + 2). We will use the quotient rule and chain rule to find the derivative. This is a common problem in calculus, and understanding how to solve it will help you with other differentiation problems.

Question: How to Find dy/dx for y = 10^x / (x + 2)?

Find dy/dx given the function y = 10^x / (x + 2).