Problem 22 - University Calculus: Early Transcendentals

Question

In Exercises \(9-22,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x\). $$y=e^{\left(4 \sqrt{x}+x^{2}\right)}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = e^{4\sqrt{x} + x^2} \left( \frac{2}{\sqrt{x}} + 2x \right) \).

Step 1: Identify the Inner and Outer Functions

The expression is given as \( y = e^{4\sqrt{x} + x^2} \). We need to decompose this into two functions: the inner function \( u \) and the outer function \( f(u) \). We can let \( u = 4\sqrt{x} + x^2 \), then \( y = e^u \).

Step 2: Differentiate the Outer Function

We have \( y = f(u) = e^u \). The derivative of the outer function with respect to \( u \) is \( f'(u) = e^u \).

Step 3: Differentiate the Inner Function

Now, differentiate the inner function \( u = 4\sqrt{x} + x^2 \) with respect to \( x \). We have, \( \frac{du}{dx} = \frac{d}{dx}(4\sqrt{x}) + \frac{d}{dx}(x^2) = 4 \cdot \frac{1}{2\sqrt{x}} + 2x = \frac{2}{\sqrt{x}} + 2x \).

Step 4: Apply the Chain Rule

To get \( \frac{dy}{dx} \), we apply the chain rule: \( \frac{dy}{dx} = f'(u) \cdot \frac{du}{dx} \). Substitute \( f'(u) = e^u \) and \( \frac{du}{dx} = \frac{2}{\sqrt{x}} + 2x \), so \( \frac{dy}{dx} = e^u \left( \frac{2}{\sqrt{x}} + 2x \right) \).

Step 5: Substitute Back for \( u \)

Replace \( u \) back with \( 4\sqrt{x} + x^2 \) in the expression for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = e^{4\sqrt{x} + x^2} \left( \frac{2}{\sqrt{x}} + 2x \right) \).

Key Concepts

Chain Rule

When finding derivatives in differential calculus, especially for functions composed of an inner and an outer function, the chain rule becomes incredibly useful. The chain rule helps us differentiate composite functions and is often stated in the simple form: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where \( y \) is a function of \( u \) and \( u \) is a function of \( x \). This technique is essential when dealing with functions like powers of exponentials or nested roots.

• To apply it, first identify your outer and inner functions.
• Differentiate each part separately before combining them.

For our example of \( y = e^{4\sqrt{x} + x^2} \), the rule tells us that the derivative of this whole expression involves multiplying the derivative of \( e^u \) by the derivative of \( u \) with respect to \( x \). This method ensures we calculate the derivative accurately even for complex expressions. Applying the chain rule step by step not only helps in finding the derivative effectively but also gives insight into the structure of the function.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable—often thought of as finding the "slope" of the function at any given point. In calculus, finding derivatives accurately is a crucial skill. For the function \( y = e^{4\sqrt{x} + x^2} \), derivatives help us analyze how fast \( y \) changes as \( x \) changes.

• The outer function \( y = e^u \) has a straightforward derivative: \( e^u \).
• The inner function, \( u = 4\sqrt{x} + x^2 \) requires separate attention. The derivative here combines the power rule and derivative of square root, calculated as \( \frac{2}{\sqrt{x}} + 2x \).

Differentiation lets us break down complex expressions like these, giving us tools to predict and analyze function behavior efficiently. The ability to determine \( \frac{dy}{dx} \) is foundational not only in mathematics but also in fields relying on rate changes, such as physics and economics.

Functions

In mathematics, functions are like machines; they take inputs, transform them, and produce outputs. Understanding functions is essential to performing tasks in calculus like differentiation. In our example, we distill the continuous expression \( y = e^{4\sqrt{x} + x^2} \) into manageable components.

• The outer function \( f(u) = e^u \) is an exponential function, ubiquitous due to its constant rate of growth.
• The inner function \( g(x) = 4\sqrt{x} + x^2 \) combines clear mathematical operations like square roots and polynomials, allowing us to express complex curves and behaviors.

By transforming our view from a single function to a blend of inner and outer functions, we not only simplify problems but also enable the application of calculus rules efficiently. This understanding of functions as both standalone entities and parts of larger expressions enhances our ability to manipulate and predict mathematical outcomes easily.