Problem 70 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=2^{\left(s^{2}\right)}$$

Step-by-Step Solution

Answer

\( \frac{dy}{ds} = 2^{s^2} \cdot 2s \ln(2) \).

Step 1: Identify the Function Structure

We start by identifying the function we are tasked to differentiate. The given function is \( y = 2^{s^2} \). This is an exponential function where the base is a constant (2) and the exponent is a function of \( s \), specifically \( s^2 \).

Step 2: Apply the Chain Rule

To differentiate this function, we'll need to use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \times g'(x) \). Here, we consider \( f(u) = 2^u \) with \( u = s^2 \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( f(u) = 2^u \) with respect to \( u \). The derivative of \( a^x \) with respect to \( x \) is \( a^x \ln(a) \). Applying this, we get \( \frac{d}{du} 2^u = 2^u \ln(2) \).

Step 4: Differentiate the Inner Function

Now, differentiate the inner function \( u = s^2 \) with respect to \( s \). The derivative of \( s^2 \) is \( 2s \).

Step 5: Combine the Derivatives

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function:\[\frac{dy}{ds} = 2^{s^2} \ln(2) \cdot 2s = 2^{s^2} \cdot \ln(2) \cdot 2s\]Thus, the derivative of \( y \) with respect to \( s \) is \( 2^{s^2} \cdot 2s \ln(2) \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. Essentially, the derivative tells us the rate at which one quantity changes relative to another. For instance, if you are traveling in a car, the speedometer shows the rate at which your distance from a starting point changes over time. This rate is essentially a derivative of the distance with respect to time. When dealing with exponential functions like the one in our problem, where the function is in the form of an exponential expression, calculating the derivative becomes more intricate. Here, specific differentiation techniques are required. We will break down these techniques in detail in the subsequent sections.

The Chain Rule Explained

The chain rule is an essential tool in differentiation. It is used to find the derivative of composite functions, those that involve one function inside another. If you have two functions, say \( f(x) \) and \( g(x) \), and you want to differentiate their composition, the chain rule comes into play.To put it simply, if you have a function \( y = f(g(x)) \), the chain rule allows you to differentiate by:

• Taking the derivative of the outer function \( f \) with respect to the inner function \( g(x) \).
• Then multiplying it by the derivative of the inner function \( g \) with respect to \( x \).

This can be expressed as: \[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]In our example, we used the chain rule to differentiate the function \( 2^{s^2} \) by identifying the outer function as \( 2^u \) and the inner function as \( u = s^2 \).

Differentiation Techniques for Exponential Functions

Exponential functions have a unique nature, particularly when the base of the exponential is a constant, like in \( y = 2^{s^2} \). To find their derivatives, a special rule is used. Usually, for an exponential function \( a^x \), the derivative with respect to \( x \) is \( a^x \ln(a) \).In the case of composite functions, once the chain rule identifies the outer and inner layers of the function, you differentiate each and then combine the results. In our problem, the outer function \( f(u) = 2^u \) was differentiated to get \( 2^u \ln(2) \). The inner function \( u = s^2 \) is differentiated to become \( 2s \).Finally, you combine these results using the principle of the chain rule, which gives: \[\frac{dy}{ds} = 2^{s^2} \cdot 2s \ln(2)\]This demonstrates the beautiful interplay of rules and techniques in calculus that allow us to effectively manage even the most complex of functions.