Finding Dy: Step-by-Step Solution For Y = (5x^2 + 2)/(5x + 2)

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Hey guys! Today, we're diving into a calculus problem where we need to find dy, also known as the differential of y, for a given function. We’ll then evaluate it at a specific point. Sounds interesting? Let's get started!

Understanding the Problem

Our mission is to find dy, which represents an infinitesimal change in y, for the function y = (5x² + 2) / (5x + 2). We're also given that we need to evaluate this when x = 4 and dx = 0.6, where dx represents an infinitesimal change in x. Basically, we're looking at how a tiny change in x affects y at a particular point.

Step 1: Finding the Derivative dy/dx

The first thing we need to do is find the derivative of y with respect to x, which is written as dy/dx. Since our function y is a quotient of two expressions, we'll use the quotient rule. Remember the quotient rule? It states that if y = u/v, then dy/dx = (v(du/dx) - u(dv/dx)) / v². This is a fundamental concept in calculus, so make sure you're comfortable with it.

In our case:

  • u = 5x² + 2
  • v = 5x + 2

Let's find the derivatives of u and v with respect to x:

  • du/dx = 10x (using the power rule)
  • dv/dx = 5 (simple derivative of a linear function)

Now, we plug these into the quotient rule:

dy/dx = ((5x + 2)(10x) - (5x² + 2)(5)) / (5x + 2)²

Let's simplify this expression. Expanding the numerator gives us:

dy/dx = (50x² + 20x - 25x² - 10) / (5x + 2)²

Combining like terms in the numerator, we get:

dy/dx = (25x² + 20x - 10) / (5x + 2)²

So, we've found the derivative dy/dx. This tells us the instantaneous rate of change of y with respect to x at any point. It's a crucial step in solving our problem.

Step 2: Expressing dy in Terms of dx

Now that we have dy/dx, we can express dy in terms of dx. This is a simple algebraic manipulation. We just multiply both sides of the equation by dx:

dy = (dy/dx) * dx

Substituting the expression we found for dy/dx, we get:

dy = ((25x² + 20x - 10) / (5x + 2)²) * dx

This equation now tells us how a small change dx affects y, represented by dy. We're getting closer to our final answer!

Step 3: Evaluating dy at x = 4 and dx = 0.6

The final step is to evaluate dy when x = 4 and dx = 0.6. This means we plug in these values into our expression for dy:

dy = ((25(4)² + 20(4) - 10) / (5(4) + 2)²) * 0.6

Let's break this down:

  • 25(4)² = 25 * 16 = 400
  • 20(4) = 80
  • 5(4) + 2 = 20 + 2 = 22

So, our equation becomes:

dy = ((400 + 80 - 10) / (22)²) * 0.6

Simplifying further:

dy = (470 / 484) * 0.6

Now, let's do the division and multiplication:

dy ≈ 0.97107 * 0.6

dy ≈ 0.5826

Therefore, when x = 4 and dx = 0.6, dy is approximately 0.5826. This means that for a small change of 0.6 in x around x = 4, the value of y changes by approximately 0.5826.

Conclusion

So, there you have it! We found dy and evaluated it for the given function and values. We used the quotient rule to find the derivative, expressed dy in terms of dx, and then plugged in our values to get the final answer. Calculus can seem daunting at first, but breaking it down step-by-step makes it much more manageable. Remember, practice makes perfect, so keep working on these types of problems! This is a common type of problem in introductory calculus courses, so mastering these steps will definitely help you out. You got this!

Key Takeaways

  • Quotient Rule: Remember the quotient rule! It's essential for finding derivatives of functions that are fractions.
  • Differentials: Understanding what dy and dx represent is crucial. They're small changes in y and x, respectively.
  • Step-by-Step Approach: Breaking down complex problems into smaller, manageable steps makes them easier to solve.
  • Practice! The more you practice, the more comfortable you'll become with calculus concepts.

Additional Tips for Understanding Differentials

To truly grasp differentials, think about them geometrically. The derivative dy/dx represents the slope of the tangent line to the curve y = f(x) at a specific point. The differential dy is an approximation of the change in y along this tangent line for a small change dx in x. This approximation is most accurate when dx is very small. Visualizing this relationship can be incredibly helpful!

Also, remember that differentials have applications beyond just finding approximate changes in functions. They are fundamental in integration, differential equations, and many other areas of mathematics and physics. So, spending the time to understand them well is a worthwhile investment in your mathematical journey.

If you are still struggling with this concept please leave a comment below and I will try to explain it further! Good luck, guys! You can solve this kind of problem, and remember practice is key to really mastering these concepts!