Problem 28 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ g(s)=\exp \left[\tan s^{3}\right] $$

Step-by-Step Solution

Answer

\( g'(s) = 3s^2 \exp(\tan(s^3)) \sec^2(s^3) \).

Step 1: Identify the Outer and Inner Functions

The function to differentiate is given as \( g(s) = \exp[\tan(s^3)] \). Here, \( \exp(u) \) is the outer function with respect to \( u = \tan(s^3) \), and \( \tan(v) \) is the second function where \( v = s^3 \).

Step 2: Differentiate the Outer Function

The derivative of the outer function \( \exp(u) \) with respect to \( u \) is \( \exp(u) \). Using the chain rule, \( \frac{d}{ds} \exp[\tan(s^3)] = \exp[\tan(s^3)] \cdot \frac{d}{ds}[\tan(s^3)] \).

Step 3: Differentiate the Inner Function – Part 1

The inner function in this chain \( \tan(s^3) \) must be differentiated with respect to \( s \). Start by differentiating the function \( \tan(w) \), where \( w = s^3 \). We have \( \frac{d}{dw} \tan(w) = \sec^2(w) \).

Step 4: Differentiate the Inner Function – Part 2

Now differentiate the innermost function \( w = s^3 \) with respect to \( s \), which is \( \frac{dw}{ds} = 3s^2 \).

Step 5: Apply the Chain Rule to the Inner Function

Combine Steps 3 and 4 using the chain rule: \( \frac{d}{ds}[\tan(s^3)] = \sec^2(s^3) \cdot 3s^2 \).

Step 6: Combine Results from Steps 2 and 5

Substitute \( \frac{d}{ds}[\tan(s^3)] = \sec^2(s^3) \cdot 3s^2 \) into the result from Step 2: \[\frac{d}{ds} [\exp(\tan(s^3))] = \exp(\tan(s^3)) \cdot \sec^2(s^3) \cdot 3s^2.\]

Step 7: Write the Final Expression

The derivative of \( g(s) = \exp[\tan(s^3)] \) with respect to \( s \) is: \[ g'(s) = 3s^2 \exp(\tan(s^3)) \sec^2(s^3).\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. These are functions composed of two or more other functions. It allows you to find the derivative of a function by breaking it down into its simpler components. When you have a function like \( g(s) = \exp[\tan(s^3)] \), you can see it consists of an exponential function and a trigonometric function combined in a complex way.
To apply the chain rule, you need to identify the inner function and the outer function:

• The outer function here is \( \exp(u) \), where \( u = \tan(s^3) \).
• The inner function is \( \tan(v) \), where \( v = s^3 \).

The chain rule helps to differentiate these nested layers one step at a time. It states that if a function \( h(x) = f(g(x)) \), then its derivative is \( h'(x) = f'(g(x)) \cdot g'(x) \). In our example, this rule allows you to connect the derivative of each layer effectively.

Exponential Function

Exponential functions are functions of the form \( \exp(u) \) or its equivalent \( e^u \), where \( e \) is the base of natural logarithms (approximately 2.718). The distinct feature of these functions is their constant growth rate, which entails that their derivative is the same as the original function itself.
When differentiating an exponential function within another function, it is crucial to consider the chain rule. For \( \exp[u] \), where \( u \) itself is a function of \( s \), the differentiation gives:

• The derivative of \( \exp(u) \) with respect to \( u \) is \( \exp(u) \).
• Thanks to the chain rule, multiply it by the derivative of \( u \) with respect to \( s \).

In our case, since \( u = \tan(s^3) \), it leads to \( \frac{d}{ds} \exp(\tan(s^3)) = \exp(\tan(s^3)) \cdot \frac{d}{ds}[\tan(s^3)] \), allowing you to further work on the inner function.

Trigonometric Differentiation

Trigonometric functions like \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are often encountered in calculus. When it comes to differentiating these, each has a specific rule. For the tangent function, which appears in our problem, you need to know:

• The derivative of \( \tan(w) \) is \( \sec^2(w) \).

Understanding this helps when differentiating more complex expressions involving trigonometric terms.
In the given function, \( \tan(s^3) \) appears, which means we need to find its derivative using \( \sec^2(s^3) \). Multiplying this by the derivative of \( s^3 \) (applying the chain rule) leads to completer the differentiation step for the trigonometric part: \( \frac{d}{ds} \tan(s^3) = \sec^2(s^3) \cdot 3s^2 \). This result is used to complete the overall differentiation task.

Derivative of Composite Functions

Composite functions are those made by combining two or more functions into a single expression. Differentiating such compositions requires careful application of the chain rule.For the function \( g(s) = \exp[\tan(s^3)] \), you break it down into multiple layers:

• The outer function is \( \exp[u] \).
• The middle function is \( \tan[v] \).
• The innermost function is \( v = s^3 \).

To find the derivative, start from the outermost layer and work inward. As derived, the steps are:1. Differentiate the outer function: \( \frac{d}{ds} \exp(\tan(s^3)) = \exp(\tan(s^3)) \cdot \frac{d}{ds}[\tan(s^3)] \).2. Use the chain rule to find the derivative of the inner function: \( \frac{d}{ds}[\tan(s^3)] = \sec^2(s^3) \cdot 3s^2 \).3. Combine these results for the complete derivative: \( g'(s) = 3s^2 \exp(\tan(s^3)) \sec^2(s^3) \).This method ensures you respect the layered structure of the function, ensuring accurate differentiation.