Problem 34 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$y=(2 x-5)^{-1}\left(x^{2}-5 x\right)^{6}$$

Step-by-Step Solution

Answer

Differentiate using product and chain rules to find \( y' = -2(2x-5)^{-2}(x^{2}-5x)^{6} + 6(x^{2}-5x)^{5} \).

Step 1: Identify the Function Parts

The function given is \( y = (2x-5)^{-1}(x^{2}-5x)^{6} \). Notice that this is a product of two functions: \( u(x) = (2x-5)^{-1} \) and \( v(x) = (x^{2}-5x)^{6} \). We will use the product rule for differentiation.

Step 2: Differentiate the First Function Using Chain Rule

Let \( u(x) = (2x-5)^{-1} \). To differentiate \( u(x) \) with respect to \( x \), recognize it as \( u(x) = (2x-5)^{-1} = (2x-5)^{-1} \), which is a composite function, so use the chain rule:\[ u'(x) = -1(2x-5)^{-2} \cdot 2 = -2(2x-5)^{-2}. \]

Step 3: Differentiate the Second Function Using Chain Rule

Let \( v(x) = (x^{2}-5x)^{6} \). Again, use the chain rule to differentiate: \[ v'(x) = 6(x^{2}-5x)^{5} \cdot (2x-5) = 6(x^2 - 5x)^5 (2x - 5). \]

Step 4: Apply the Product Rule

The product rule states that if \( y = u(x)v(x) \), then: \[ y' = u'(x)v(x) + u(x)v'(x). \]Substitute the differentiated components:\[ y' = [-2(2x-5)^{-2}] (x^{2}-5x)^{6} + (2x-5)^{-1} [6(x^2 - 5x)^5 (2x - 5)]. \]

Step 5: Simplify the Expression

First part of the solution:\[ -2(2x-5)^{-2}(x^{2}-5x)^{6}. \]Second part:\[ (2x-5)^{-1}(6(x^2 - 5x)^5 (2x - 5)). \]Combine both parts:\[ y' = -2(2x-5)^{-2}(x^{2}-5x)^{6} + 6(x^{2}-5x)^{5}. \]

Key Concepts

Product Rule in Differentiation

The product rule is an essential technique in calculus for finding the derivative of a product of two functions. It's particularly useful when you have complex functions multiplied together, like in our exercise where the function is given as \(y = (2x-5)^{-1}(x^{2}-5x)^{6}\). The product rule states that if \(y = u(x)v(x)\), then its derivative \(y'\) is given by the formula:

• \(y' = u'(x)v(x) + u(x)v'(x)\)

This formula tells us to differentiate each function separately (while treating the other as a constant), and then sum up the two resulting products. In our solution, both parts of the function were broken down as \(u(x) = (2x-5)^{-1}\) and \(v(x) = (x^2 - 5x)^6\).
Using the product rule, we derived the solution by not only applying this rule, but also by making use of other rules as we'll discuss next, ensuring that every aspect of the product's differentiation is captured correctly.

Understanding the Chain Rule

The chain rule is fundamental when dealing with composite functions. These are functions within functions, like \(u(x) = (2x-5)^{-1}\) and \(v(x) = (x^{2}-5x)^{6}\) in our exercise. The chain rule helps us find the derivative of such nested functions efficiently.
To use the chain rule, one begins by identifying the outer and the inner functions. For example, with the function \(u(x)\), recognize it as a power expression where the outer function is \(f(z) = z^{-1}\) and the inner function is \(z(x) = 2x-5\). Applying the chain rule means taking the derivative of the outer function and multiplying it by the derivative of the inner one:

• \(u'(x) = f'(z) \cdot z'(x) = -2(2x-5)^{-2}\)

For \(v(x)\), which is another composite function, use the same principle by identifying \(g(w) = w^{6}\) with \(w(x) = x^{2} - 5x\). This gives:

• \(v'(x) = 6(x^2 - 5x)^5 \cdot (2x - 5)\)

By understanding the chain rule, students can easily untangle layers of nested functions and smoothly compute derivatives.

Differentiation Techniques: Combined Approach

Differentiation techniques encompass a variety of methods for finding derivatives, tailored to the structure of the functions at hand. In the given exercise, mastering the product rule and the chain rule was crucial due to the nature of the function \(y = (2x-5)^{-1}(x^{2}-5x)^{6}\). Combining these rules gives us a robust toolkit to handle intricate equations.
First, identifying different parts of complex functions allows the targeted use of appropriate rules. Here's a summary of the steps taken:

• Applying the product rule required separate differentiation of functions \(u(x)\) and \(v(x)\).
• The chain rule facilitated the differentiation of the nested parts within \(u(x)\) and \(v(x)\).

Once these rules are employed, it's important to work through the algebraic simplification of the resulting expression, as shown in our final step where the parts were combined and simplified back into a single function. By learning how to effectively combine these rules, students will enhance their problem-solving skills and gain greater confidence in tackling complex calculus problems.