Problem 5 - Thomas Calculus

Question

Find \(d y / d x\) in Exercises \(1-10\) $$ y=7 \sqrt{x+6} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{7}{2\sqrt{x+6}} \).

Step 1: Identify the Function

The function we need to differentiate is given as \( y = 7 \sqrt{x+6} \). Our goal is to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \).

Step 2: Rewrite the Square Root

Instead of using the square root, rewrite the function using exponent notation. The function becomes \( y = 7 (x+6)^{1/2} \).

Step 3: Apply the Power Rule

Use the power rule of differentiation which states that if \( y = ax^n \), then \( \frac{dy}{dx} = a \cdot n \cdot x^{n-1} \). Here, let \( u = x+6 \) and apply the power rule: \( \frac{d}{dx}[u^{1/2}] = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx} \).

Step 4: Differentiate the Inner Function

Differentiate \( u = x+6 \) with respect to \( x \). This gives us \( \frac{du}{dx} = 1 \).

Step 5: Combine Derivatives

Substituting back the terms derived, we have \( \frac{dy}{dx} = 7 \cdot \frac{1}{2} \cdot (x+6)^{-1/2} \cdot 1 \). Simplify to get \( \frac{dy}{dx} = \frac{7}{2} (x+6)^{-1/2} \).

Step 6: Finalize the Derivative Expression

Convert the expression back into a square root form: \( \frac{dy}{dx} = \frac{7}{2 \cdot \sqrt{x+6}} \). This is the simplified version of the derivative.

Key Concepts

Derivative of a Function

The derivative of a function is a fundamental concept in calculus. It measures how a function's output value, or the dependent variable, changes with respect to its input value, or the independent variable. In simpler terms, it's like measuring how fast something is moving or growing at any point in time. This is often used to find rates of change, tangents to curves, and optimize functions.

In mathematical notation, the derivative of a function \( y \) with respect to \( x \) is denoted by \( \frac{dy}{dx} \). This represents the slope of the tangent line to the function's graph at any given point. To find this, we use the rules of differentiation, which guide us on how to manipulate functions to obtain their derivatives.

• Derivatives provide a way to predict the behavior of dynamic systems in fields like physics, engineering, economics, and more.
• Understanding how to compute derivatives is key to solving more complex calculus problems.

Power Rule

The power rule is one of the simplest and most essential rules for differentiation. It states that if you have a function of the form \( y = ax^n \), where \( a \) is a constant and \( n \) is a real number, the derivative is \( \frac{dy}{dx} = a \cdot n \cdot x^{n-1} \).

This rule helps turn the process of differentiation into a straightforward algorithmic step, requiring you to only multiply and subtract.

• You multiply the entire expression by the power of \( x \) (\( n \)).
• Then, decrease the power of \( x \) by one to get the new exponent.

The power rule is particularly useful when dealing with polynomial functions. It simplifies the task of finding derivatives and makes handling complex expressions more manageable.

Square Root Function

Square root functions are a type of radical function defined as the principal square root of \( x \), often written as \( \sqrt{x} \). These functions are typically expressed in the form \( y = a\sqrt{x + b} \) where \( a \) and \( b \) are constants.

When differentiating square root functions, it's advantageous to use exponent notation to rewrite the function, as this form is more compatible with the power rule. For instance, \( \sqrt{x} \) is equivalent to \( x^{1/2} \).

• Rewriting square roots with exponents allows the application of the power rule for easier differentiation.
• This method transforms the function into a polynomial-like form, which is simple to handle using existing differentiation rules.

Understanding how to manipulate square root functions is crucial not only for derivatives but also for integrals and solving equations.