91Ó°ÊÓ

Differentiation is a fundamental concept in calculus, and it's all about understanding how a function changes. When you differentiate a function, you're essentially finding its derivative. This represents the rate at which the function's output value changes as its input changes. Let's break this down further.
When we look at the function \(y = x \sqrt{1 + 4x}\), differentiation gives us a formula to predict how \(y\) changes with \(x\). For this, we apply the product rule, which is vital when dealing with products of functions. The product rule states that for any two functions \(u\) and \(v\),

• \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\)

Here's how it was done in the solution:

• \(u = x\) and \(v = \sqrt{1 + 4x}\)
• \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{2}{\sqrt{1 + 4x}}\)
• The derivative \(\frac{dy}{dx} = \frac{2x}{\sqrt{1 + 4x}} + \sqrt{1 + 4x}\)

By substituting \(x = 12\) into this derivative, we find how \(y\) changes at \(x = 12\). Calculating this gives the slope of the tangent, a crucial part of understanding the behavior of functions.

In calculus, approximation is a handy tool, especially when dealing with small changes. Consider Taylor series or linear approximations, both of which offer ways to estimate function values at points where they aren't directly calculated.
Approximations come into play with the concept of differentials. If you have a function that's hard to evaluate directly, a small change \(\Delta x\) can be approximated to find \(\Delta y\). However, it's often more useful to work with the differential \(dy\), which is calculated as \(dy = \frac{dy}{dx} \cdot \Delta x\). This shows how \(dy\) acts as a linear estimate of \(\Delta y\), especially when \(\Delta x\) is small.
When \(\Delta x = 0.06\) and the derivative at \(x = 12\) is \(\frac{73}{7}\), the calculation \(dy = \frac{73}{7} \cdot 0.06\) gives us approximately \(0.6268\). This serves as an estimation for the actual change \(\Delta y\), showing how calculus provides a close and useful approximation for small intervals.

One of the most significant applications of differentiation is to find the rate of change. This tells how one quantity changes in relation to another, and is often represented as the derivative \(\frac{dy}{dx}\).
In our example with the function \(y = x \sqrt{1 + 4x}\), finding \(\frac{dy}{dx}\) helps us understand how the function's output value changes as we change \(x\). The computed derivative represents this rate of change, showing how sensitive the function is to variations in \(x\).
The solution calculates this rate for \(x = 12\) by substituting into the derived formula. This yields \(\frac{dy}{dx} = \frac{73}{7}\). Such information is vital in fields like physics and engineering, where understanding how quickly something changes is crucial for prediction and control. For instance, knowing the rate at which a vehicle moves allows for estimations of distance over time or to strategize better maneuvers.