Differentiating Functions A Step-by-Step Guide To Solving Calculus Problems

Hey guys! Today, we're diving deep into the world of differentiation, a fundamental concept in calculus. We're going to break down three different functions and learn how to find their derivatives. Differentiation, at its core, is about finding the instantaneous rate of change of a function. Think of it as zooming in on a curve until it looks like a straight line and then finding the slope of that line. This slope tells us how the function is changing at that specific point. So, grab your calculators and let's get started!

1. Differentiating y = \frac{15}{x^5} + \frac{\sqrt[3]{x^2}}{2} + K

Let's start with our first function: y = 15/x⁵ + (x^(⅔))/2 + K. This might look a bit intimidating at first, but don't worry, we'll tackle it step by step. The key here is to remember the power rule and how to handle constants. The power rule states that if we have a term like xⁿ, its derivative is n x^(n-1). In simpler terms, we multiply by the exponent and then reduce the exponent by 1. Also, remember that the derivative of a constant (like K) is always zero. This is because a constant doesn't change, so its rate of change is zero.

Before we can directly apply the power rule, we need to rewrite our function to make the exponents clearer. We can rewrite 15/x⁵ as 15x⁻⁵ and ∛(x²) as x^(⅔). This makes it much easier to see the exponents we'll be working with. So, our function now looks like this: y = 15x⁻⁵ + (1/2)x^(⅔) + K. Now we are ready to differentiate. Applying the power rule to the first term, 15x⁻⁵, we multiply by the exponent (-5) and reduce the exponent by 1, giving us -75x⁻⁶. For the second term, (1/2)x^(⅔), we multiply by the exponent (⅔) and reduce the exponent by 1, resulting in (⅓)x^(-⅓). And finally, the derivative of the constant K is simply 0. Therefore, combining these results, we get the derivative dy/dx = -75x⁻⁶ + (⅓)x^(-⅓). But we're not quite done yet! It's good practice to rewrite our answer without negative exponents. Remember that x⁻ⁿ is the same as 1/xⁿ. So, we can rewrite our derivative as dy/dx = -75/x⁶ + 1/(3∛x). And there you have it! We've successfully differentiated the first function.

In this function, we encountered a combination of power rule application and dealing with constants. The transformation of the function into a more manageable form using negative and fractional exponents was crucial. Understanding these manipulations is key to successfully differentiating various types of functions. Furthermore, simplifying the final result by removing negative exponents presents the derivative in its most conventional form, making it easier to interpret and use in further calculations or applications. By mastering this process, you'll be well-equipped to handle more complex differentiation problems in the future. Remember, practice makes perfect, so the more you work through these types of problems, the more comfortable you'll become with the process.

2. Differentiating y = \frac{x-4}{x-2}

Next up, we have y = (x-4)/(x-2). This function is a bit different because it's a quotient – we're dividing one expression by another. For this, we need to use the quotient rule. The quotient rule states that if we have a function y = u/v, where u and v are both functions of x, then the derivative dy/dx is given by (v(du/dx) - u(dv/dx)) / v². In simpler terms, it's (bottom times derivative of the top minus top times derivative of the bottom) all over the bottom squared. It might sound a bit complicated, but it's a straightforward rule once you get the hang of it.

So, let's identify our u and v. In this case, u = x - 4 and v = x - 2. Now we need to find the derivatives of u and v. The derivative of u = x - 4 is simply 1, since the derivative of x is 1 and the derivative of a constant (-4) is 0. Similarly, the derivative of v = x - 2 is also 1. Now we have all the pieces we need to apply the quotient rule. Plugging everything into the formula, we get: dy/dx = ((x-2)(1) - (x-4)(1)) / (x-2)². Now we need to simplify this expression. Expanding the numerator, we get dy/dx = (x - 2 - x + 4) / (x-2)². Notice that the x terms cancel out, leaving us with dy/dx = 2 / (x-2)². And that's our derivative! We've successfully used the quotient rule to differentiate this function.

This example highlights the importance of recognizing function structures and applying the appropriate differentiation rules. The quotient rule is a powerful tool for handling functions that are expressed as a ratio. Moreover, the algebraic simplification step is equally crucial, as it condenses the derivative into its simplest form, which is easier to interpret and apply in further calculations. In practical applications, derivatives often represent rates of change, and a simplified form allows for easier analysis and decision-making based on these rates. Therefore, mastering the quotient rule, along with the ability to simplify expressions, forms a cornerstone of differential calculus.

3. Differentiating y = x² - 2x + 1

Finally, let's look at our third function: y = x² - 2x + 1. This is a polynomial function, and differentiating it is relatively straightforward. We'll be using the power rule again, along with the fact that the derivative of a constant multiplied by a function is just the constant times the derivative of the function. Remember, the power rule states that if we have a term like xⁿ, its derivative is n x^(n-1).

Let's differentiate each term separately. For the first term, x², we apply the power rule: multiply by the exponent (2) and reduce the exponent by 1, giving us 2x. For the second term, -2x, we again apply the power rule (remember that x is the same as x¹): multiply by the exponent (1) and reduce the exponent by 1, giving us -2. The derivative of the constant term, 1, is 0. Therefore, combining these results, we get the derivative dy/dx = 2x - 2. We've successfully differentiated this polynomial function!

This final example reinforces the application of the power rule, which is a cornerstone technique in differential calculus. Differentiating polynomials is a foundational skill, and understanding how to apply the power rule efficiently is essential for tackling more complex functions. The simplicity of this example allows for a clear understanding of how each term contributes to the final derivative, highlighting the additive nature of differentiation. Moreover, the ability to quickly differentiate polynomial functions is invaluable in various applications, including optimization problems, curve sketching, and approximation methods. By mastering this fundamental skill, you'll build a strong base for further exploration in calculus and its applications.

Conclusion

So, guys, we've successfully differentiated three different functions today! We covered the power rule, the quotient rule, and how to handle constants. Remember, the key to mastering differentiation is practice. The more you work through problems, the more comfortable you'll become with the rules and techniques. Differentiation is a powerful tool with applications in physics, engineering, economics, and many other fields. Keep practicing, and you'll be differentiating like a pro in no time!