Derivative Of Y = E^x Log X Calculation And Explanation
Hey guys! Today, we're diving into a super interesting calculus problem that involves finding the derivative of a function that combines both an exponential and a logarithmic term. Specifically, we're tackling the function y = e^x * log(x). This is a classic example that tests your understanding of the product rule and the derivatives of exponential and logarithmic functions. So, buckle up, and let's get started!
Understanding the Function: y = e^x * log(x)
Before we jump into finding the derivative, let's break down what this function actually represents. We have y as a function of x, where y is the product of two distinct functions: e^x (the exponential function) and log(x) (the natural logarithm function). The exponential function, e^x, is one of the most fundamental functions in calculus, known for its unique property that its derivative is itself. The natural logarithm function, log(x), on the other hand, is the inverse of the exponential function and represents the power to which the base e must be raised to obtain x. Understanding how these two functions behave individually is crucial before we can understand how they interact when multiplied together.
When we multiply these two functions, we get a new function, y = e^x * log(x), which exhibits a more complex behavior. The exponential term e^x will cause the function to grow rapidly as x increases, while the logarithmic term log(x) will moderate this growth, especially for smaller values of x. This interplay between exponential growth and logarithmic moderation is what makes this function interesting and also what makes finding its derivative a worthwhile exercise. Think of it like this: e^x is the accelerator in a car, constantly pushing the speed higher, while log(x) is like the brakes, gently keeping things in check. The resulting speed is a balance between these two forces.
Now, why are we even interested in the derivative of this function? Well, the derivative, dy/dx, tells us the instantaneous rate of change of y with respect to x. In simpler terms, it tells us how much y is changing for a tiny change in x. This information is incredibly valuable in many applications, such as optimization problems (finding maximum or minimum values), curve sketching (understanding the shape of the graph), and physics (calculating velocity and acceleration). So, finding dy/dx for y = e^x * log(x) will give us insights into how this combined function changes and behaves.
Applying the Product Rule
The heart of solving this problem lies in applying the product rule. The product rule is a fundamental concept in differential calculus that provides a formula for finding the derivative of a product of two functions. If we have two functions, say u(x) and v(x), then the derivative of their product, d/dx [u(x)v(x)], is given by: u'(x)v(x) + u(x)v'(x). In simpler words, it states that the derivative of the product is the derivative of the first function times the second function plus the first function times the derivative of the second function. This rule is absolutely essential when dealing with functions that are formed by multiplying two other functions together.
In our case, we have y = e^x * log(x), which can be seen as the product of two functions: u(x) = e^x and v(x) = log(x). To apply the product rule, we first need to find the derivatives of these individual functions. The derivative of e^x is simply e^x, which is one of the most beautiful results in calculus. The derivative of log(x), on the other hand, is 1/x. These are standard derivatives that you should have memorized or at least be very familiar with. Knowing these basic derivatives is like knowing your times tables in arithmetic – it makes everything else much easier.
Now, we have all the pieces we need to apply the product rule. Let's plug our functions and their derivatives into the formula: dy/dx = u'(x)v(x) + u(x)v'(x). Substituting u(x) = e^x, v(x) = log(x), u'(x) = e^x, and v'(x) = 1/x, we get: dy/dx = e^x * log(x) + e^x * (1/x). This expression represents the derivative of our function y = e^x * log(x). It's a combination of the original functions and their derivatives, showing how the rate of change of y depends on both the exponential and logarithmic terms. This is the power of the product rule – it allows us to break down complex derivatives into simpler components.
Simplifying the Result
Okay, we've successfully applied the product rule and found the derivative, but the job isn't quite done yet. The next step is to simplify the expression we obtained to make it more readable and easier to work with. This is a crucial step in many calculus problems because a simplified answer is often more insightful and less prone to errors in further calculations. Think of simplifying as tidying up your work – it makes everything clearer and more manageable.
Our current expression for the derivative is: dy/dx = e^x * log(x) + e^x * (1/x). Notice that both terms have a common factor of e^x. This suggests that we can factor out e^x to simplify the expression. Factoring out e^x, we get: dy/dx = e^x [log(x) + (1/x)]. This simplified expression is much cleaner and more compact than the original. It clearly shows the contribution of the exponential term e^x and the combined logarithmic and reciprocal terms within the brackets.
We can also rewrite the expression inside the brackets to make it even more visually appealing. Instead of writing log(x) + (1/x), we can write (1/x) + log(x). The order of addition doesn't change the value, but writing it this way aligns with the common notation of placing simpler terms (like 1/x) before more complex terms (like log(x)). So, our final simplified expression for the derivative is: dy/dx = e^x [(1/x) + log(x)]. This is the derivative we were looking for, and it matches one of the answer choices provided in the original problem. Simplifying not only makes the answer look better, but it also makes it easier to recognize the correct option among multiple choices.
Matching with the Options
Now that we have our simplified derivative, dy/dx = e^x [(1/x) + log(x)], the final step is to match it with the given options. This is a crucial part of problem-solving because it ensures that we haven't made any mistakes along the way and that our answer is indeed the correct one. It's like double-checking your work before submitting a final report – it's always a good idea to be sure.
Looking back at the original problem, we had four options to choose from: a) (1/e^x)(x + log x), b) e^x(1/x + log x), c) e^x(x + 1/log x), and d) (1/e^x)(1/x + log x). Comparing our simplified derivative, dy/dx = e^x [(1/x) + log(x)], with these options, we can clearly see that it matches option b) exactly. This is a satisfying moment when everything comes together and you know you've solved the problem correctly.
Option a) is incorrect because it has (1/e^x) instead of e^x and a different expression inside the brackets. Option c) is incorrect because it has (x + 1/log x) inside the brackets, which is not the same as (1/x + log x). Option d) is also incorrect because it has (1/e^x) instead of e^x. Therefore, by carefully comparing our derived expression with the given options, we can confidently conclude that option b) is the correct answer. This process of elimination and comparison is a valuable skill in any problem-solving scenario.
Conclusion
So, there you have it! We've successfully found the derivative of y = e^x * log(x) using the product rule and simplified the result to match one of the given options. This problem is a great example of how to apply fundamental calculus concepts to solve more complex problems. Remember, the key is to break down the problem into smaller, manageable steps, apply the appropriate rules and formulas, and simplify your answer as much as possible. Think of it like building a house – you start with the foundation (basic concepts), then add the walls and roof (rules and formulas), and finally, you decorate and furnish (simplify).
We started by understanding the function and recognizing that it's a product of two simpler functions, e^x and log(x). Then, we applied the product rule, which is essential for finding the derivative of a product of functions. We carefully calculated the derivatives of e^x and log(x) and plugged them into the product rule formula. After obtaining the initial derivative, we simplified the expression by factoring out the common term e^x. Finally, we compared our simplified derivative with the given options and identified the correct answer. Each step in the process is important, and mastering these steps will make you a confident problem-solver.
I hope this explanation was helpful and that you now have a better understanding of how to find the derivative of functions like y = e^x * log(x). Keep practicing, and you'll become a calculus whiz in no time! Remember, calculus is like a muscle – the more you use it, the stronger it gets. So, keep flexing those calculus muscles!
