91Ó°ÊÓ

The product rule is an essential tool when differentiating products of two functions. Imagine you have two functions, \( u(x) \) and \( v(x) \), that are multiplied together. The product rule helps us find the derivative of their product. This rule is expressed as:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function \( u \), multiply it by the second function \( v \), and then add the product of the first function \( u \) with the derivative of the second function \( v \).
In our exercise, \( (2x - 5)^{-1} \) is \( u(x) \), and \( (x^2 - 5x)^6 \) is \( v(x) \).
Applying the product rule to these functions, we'll need their respective derivatives \( u'(x) \) and \( v'(x) \), and we'll substitute these back into the formula to find the derivative of the entire function.
Once you master the product rule, it becomes much easier to handle complex differentiation problems involving multiplication of functions.

The power rule is probably the most straightforward differentiation rule to remember. This rule applies to functions where a variable is raised to a power, like \( x^n \).
To differentiate such a function, you multiply by the power, then subtract one from that power. In mathematical terms,

• \( \frac{d}{dx} x^n = nx^{n-1} \)

In the provided solution, this rule applies when we differentiate the function \( (2x-5)^{-1} \). Think of it as \( (2x-5)^n \) where \( n = -1 \).
To differentiate, bring the power \(-1\) in front and subtract one from \(-1\), resulting in \( u'(x) = -1(2x-5)^{-2} \).
The power rule simplifies many calculus problems by offering a quick way to deal with powers in differentiation.

The chain rule is invaluable when dealing with composite functions. These are functions inside other functions, such as \( (x^2 - 5x)^6 \). The chain rule allows us to differentiate such nested compositions of functions.
The basic formula of the chain rule is:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

This means you take the derivative of the outer function, evaluate it at the inner function \( g(x) \), and multiply it by the derivative of the inner function.
In our exercise, \( v(x) = (x^2 - 5x)^6 \) is a perfect example. Letting \( z = x^2 - 5x \), we identify \( v(x) = z^6 \).
Differentiating gives \( 6z^5 \). Multiply this by the derivative of \( z = x^2 - 5x \), which is \( 2x - 5 \), to find \( v'(x) = 6(x^2 - 5x)^5 (2x - 5) \).
Mastering the chain rule is essential for any calculus student, as it expands your ability to work with more intricate functions effectively.