Problem 24 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\exp (\sqrt{1+5 x^{3}})$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = e^{\sqrt{1 + 5x^3}} \cdot \frac{15x^2}{2\sqrt{1 + 5x^3}} \).

Step 1: Identify the Need for the Chain Rule

The function given is a composition of two functions: the exponential function and the square root function. To differentiate these, we'll need to apply the chain rule, which is used when differentiating composite functions.

Step 2: Let the Inner Function

Define the inner function as \( u = \sqrt{1 + 5x^3} \). The task now is to differentiate \( y = e^u \) with respect to \( u \), and also differentiate \( u \) with respect to \( x \).

Step 3: Differentiate the Outer Function with Respect to u

The outer function is \( y = e^u \). The derivative of an exponential function is the function itself, so \( \frac{dy}{du} = e^u \).

Step 4: Differentiate the Inner Function with Respect to x

The inner function is \( u = \sqrt{1 + 5x^3} \). We can rewrite this as \( u = (1 + 5x^3)^{1/2} \). Using the chain rule and the power rule, we find: \[ \frac{du}{dx} = \frac{1}{2}(1 + 5x^3)^{-1/2} \cdot 15x^2 \].

Step 5: Apply the Chain Rule to Find dy/dx

According to the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the expressions from Steps 3 and 4: \[ \frac{dy}{dx} = e^u \cdot \frac{1}{2}(1 + 5x^3)^{-1/2} \cdot 15x^2 \].

Step 6: Substitute Back the Inner Function

Substitute back \( u = \sqrt{1 + 5x^3} \) into the derivative. This gives: \[ \frac{dy}{dx} = e^{\sqrt{1 + 5x^3}} \cdot \frac{15x^2}{2\sqrt{1 + 5x^3}} \].

Key Concepts

Chain Rule

In calculus, the chain rule is vital when dealing with composite functions. A composite function is a function made by combining two or more functions. Think of it as a chain where each link is a function.
To use the chain rule, we first identify the inner and outer functions. The rule tells us to differentiate from the outside in, step by step.
Consider: if we have a function like \( y = f(g(x)) \), where \( g(x) \) is inside \( f \), the chain rule states:

• First, find the derivative of the outer function with respect to the inner function, \( \frac{dy}{du} \).
• Then, find the derivative of the inner function with respect to \( x \), \( \frac{du}{dx} \).
• Finally, multiply these derivatives together to get \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

This process allows us to manage more complex derivatives effectively, as seen in the exercise where the exponential and square root functions are intertwined.

Derivative of Exponential Functions

Exponential functions, such as \( e^x \), have unique properties. One of their most important features is their derivative.
For any exponential function of the form \( f(x) = e^{u} \), the derivative is surprisingly simple: it is the same as the function itself. This means:

• \( \frac{d}{dx}[e^u] = e^u \cdot \frac{du}{dx} \)

This property holds true for all exponential functions with the base \( e \), and helps simplify the differentiation of complex expressions quickly. In our example, the outside function \( e^u \) differentiates to \( e^u \) because of its repetitive nature. When combined with the chain rule, it makes differentiating expressions like \( y=\exp(\sqrt{1+5x^3}) \) manageable.

Composite Functions

Composite functions combine two or more functions into a single expression. This often involves placing one function inside another, such as \( f(g(x)) \).
Composite functions require specific strategies for differentiation, primarily the chain rule.
Imagine the inner function \( g(x) \) and the outer function \( f(x) \). You essentially peel back the layers, similar to an onion, starting from the outermost layer.
In the given exercise, you c identify:

• The inner function, \( u = \sqrt{1+5x^3} \), which represents the composition of a polynomial and a square root function.
• The outer function, \( y = e^u \), representing an exponential function with base \( e \).

By parsing out these elements, and applying the chain rule, differentiating complex expressions becomes a matter of simple calculation. This method helps clarify and simplify the differentiation process for students. It's a crucial skill enabling deeper understanding of calculus.