Problem 19 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=e^{\sin \left(x^{2}-1\right)} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 2x e^{\sin(x^2 - 1)} \cos(x^2 - 1) \).

Step 1: Identify the Chain Rule

We need to use the chain rule since the function is a composition of functions. The chain rule is stated as: if you have a function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). For \( f(x)=e^{ ext{something}} \), the derivative is \( e^{ ext{something}} \cdot ( ext{derivative of something}) \).

Step 2: Apply the Outer Function's Derivative

The outer function in our problem is \( e^{ ext{something}} \). The derivative of \( e^{u} \) with respect to \( u \) is \( e^{u} \). Therefore, the derivative of \( e^{ ext{something}} \) is \( e^{ ext{something}} \cdot ( ext{derivative of something}) \).

Step 3: Differentiate the Inner Function

The inner function here is \( ext{something} = \sin(x^2 - 1) \). The derivative of \( \sin(u) \) is \( \cos(u) \) and then multiply by the derivative of \( u \). So, first find the derivative of \( \sin(x^2 - 1) \), which is \( \cos(x^2 - 1) \cdot (\text{derivative of } (x^2 - 1)) \).

Step 4: Differentiate \( (x^2 - 1) \)

Now find the derivative of the innermost function, \( x^2 - 1 \). The derivative of \( x^2 \) is \( 2x \) and the derivative of \( -1 \) is \( 0 \). Thus, the derivative of \( x^2 - 1 \) is \( 2x \).

Step 5: Combine the Derivatives using the Chain Rule

Now, apply the results from Steps 2, 3, and 4 into the chain rule. The derivative of \( f(x) = e^{\sin(x^2 - 1)} \) is: \[ f'(x) = e^{\sin(x^2 - 1)} \cdot \cos(x^2 - 1) \cdot 2x \] So, the complete derivative function is: \[ f'(x) = 2x e^{\sin(x^2 - 1)} \cos(x^2 - 1) \]

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. It enables us to differentiate complex functions by breaking them down into simpler parts. When you have a composite function like \( f(g(x)) \), the chain rule specifies that the derivative \( f'(g(x)) \) is multiplied by the derivative \( g'(x) \). In terms of structure:

• Identify the outer and inner functions in the composite.
• Differentiate the outer function, leaving the inner function unchanged.
• Multiply by the derivative of the inner function.

This rule is powerful, allowing you to handle almost any function, no matter how "nested" it may be.

Composite Functions

A composite function is created when you apply one function to the results of another function. This nesting of functions can seem daunting, but breaking them down simplifies the task. For example, in the function \( f(x) = e^{\sin(x^2 - 1)} \), \( e^{\sin(x^2 - 1)} \) is the composition of the exponential function and the trigonometric function \( \sin \).

• The innermost function here is \( x^2 - 1 \).
• This result then becomes the input for the trigonometric function \( \sin(\cdot) \).
• The result of that becomes the input for the exponential function \( e^{\cdot} \).

By dissecting the function into layers, composite functions become easier to differentiate using the chain rule.

Derivative of Exponential Function

The exponential function \( e^x \) has the unique property that its derivative is the same as the original function \( e^x \). This property stays true even when dealing with composite functions. In the expression \( e^u \), where \( u = \sin(x^2 - 1) \) in our example, the derivative will remain \( e^u \) but multiplied by the derivative of \( u \) itself. Here’s how it works:

• Differentiating \( e^u \) gives \( e^u \).
• Don't forget to multiply by \( u' \) (the derivative of the inner function).

This fascinating trait of exponential functions simplifies differentiation, especially when managing functions as in our given exercise.

Trigonometric Differentiation

Differentiating trigonometric functions is a key part of many calculus problems. Each trigonometric function has its own derivative rule, such as:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• When the function is more complex, like \( \sin(x^2 - 1) \), apply the chain rule.

For \( \sin(x^2 - 1) \), first take the derivative \( \cos(x^2 - 1) \) and multiply by the derivative of \( x^2 - 1 \), which is \( 2x \). This showcases how trigonometric differentiation and the chain rule work hand-in-hand, reflecting their synergy in calculus.