Problem 86 - Calculus and its applications

Question

Differentiate. $$ y=\sqrt{(2 x-3)^{2}+1} $$

Step-by-Step Solution

Answer

\[ y' = \frac{2(2x-3)}{\sqrt{(2x-3)^2 + 1}} \]

Step 1: Identify the Outer Function

The function given is \( y=\sqrt{(2 x-3)^{2}+1} \). The outer function is the square root function, which can be rewritten as \( y = ((2x-3)^2 + 1)^{1/2} \).

Step 2: Differentiate the Outer Function Using the Chain Rule

The chain rule states that \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = u^{1/2} \), so \( f'(u) = \frac{1}{2}u^{-1/2} \). The derivative of the outer function is \( \frac{1}{2}((2x-3)^2 + 1)^{-1/2} \).

Step 3: Identify the Inner Function

The inner function is \( g(x) = (2x-3)^2 + 1 \).

Step 4: Differentiate the Inner Function

Differentiate \( g(x) = (2x-3)^2 + 1 \). Using the power rule and chain rule, the derivative is \( g'(x) = 2 \cdot (2x-3) \cdot 2 = 4(2x-3) \).

Step 5: Apply the Chain Rule to Obtain the Overall Derivative

According to the chain rule, the derivative of \( y \) with respect to \( x \) is: \[ y' = \frac{1}{2}((2x-3)^2 + 1)^{-1/2} \cdot 4(2x-3) \]Simplify this to obtain: \[ y' = \frac{2(2x-3)}{\sqrt{(2x-3)^2 + 1}} \]

Key Concepts

Using the Chain Rule in Calculus

The Chain Rule is a fundamental concept in calculus used when dealing with the derivative of a composed function, that is, a function made up of other functions. When you have a function, say \( f(g(x)) \), applying the Chain Rule allows you to differentiate it efficiently. Think of it this way: if a function is wrapped inside another, to differentiate it, you first unwrap the outer layer, then the inner layer, like peeling an onion.
For instance, consider an example where the outer function is \( y = u^{1/2} \) and the inner function is \( u = g(x) \). By using the Chain Rule, the derivative \( \frac{d}{dx}[f(g(x))] \) becomes \( f'(g(x)) \cdot g'(x) \). This means you firstly differentiate the outer function while keeping the inside unchanged. You then multiply by the derivative of the inner function.
This efficient technique helps in differentiating complex composed functions without getting lost in calculations, making it a must-know tool in calculus.

Steps in Calculating Derivatives

Derivative Calculation is a key task in calculus. To compute a derivative means to find the rate at which a function changes at any point. This involves following a series of steps that guide you through the breaking down of more complex functions.
Here's a breakdown:

• Identify the structure of the function: Is it a product, quotient, or composition of functions?
• Apply the relevant rule for the type of function: Choose between power rule, product rule, quotient rule, or chain rule.
• Find the derivative of each component: Isolate small parts, like single terms or simpler functions, and find their derivatives.
• Reassemble the parts: Combine the parts using the operation rules, such as adding, subtracting, multiplying, etc.

In our exercise, differentiating the function involved the use of both the chain rule and power rule successively to correctly handle the composed structure of the function. This calculated derivative is essential in understanding how our original function behaves as \( x \) varies.

Differentiation of Functions

Functions Differentiation is the process of finding a function’s derivative, which provides various information about the function’s behavior and is crucial in fields like physics, engineering, and economics.
Differentiation can reveal details about the slope of a curve, the rate of change, or even predict future values. To differentiate a function, mathematicians use rules based on function types, such as polynomial, trigonometric, or exponential functions.

• Every function has its specifics: For instance, polynomials use the power rule for differentiation.
• Application of rules simplifies complex differentiation: For example, the chain rule is used for nested functions, offering an easier path to the derivative.

The solution to the exercise above illustrates how differentiation of both the inner and outer functions provides insights into the overall derivative of the composed function, giving a detailed view of its change.