Problem 72 - Calculus for Biology and Medicine

Question

Find the derivative of $$ f(x)=\sec ^{2}\left(2 x^{2}-2\right) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \sec^2(2x^2 - 2) \) is \( 8x \sec^2(2x^2 - 2) \tan(2x^2 - 2) \).

Step 1: Identify the components

The function given is \[ f(x) = \sec^{2}(2x^{2} - 2) \]. It consists of an outer function, which is \( u^2 \) where \( u = \sec(2x^2 - 2) \), and an inner function \( v(x) = 2x^2 - 2 \). Use the chain rule to differentiate.

Step 2: Differentiate the outer function

The outer function is \( u^2 \), so its derivative with respect to \( u \) is \( 2u \). Since \( u = \sec(v(x)) \), we express it as \( 2 \sec(v(x)) \).

Step 3: Differentiate the middle function

The middle function is \( \sec(v(x)) \). The derivative of \( \sec(x) \)is \( \sec(x)\tan(x) \). So, the derivative of \( \sec(v(x)) \)is \( \sec(v(x))\tan(v(x)) \).

Step 4: Differentiate the inner function

Now differentiate the inner function, \( v(x) = 2x^2 - 2 \).The derivative is \( \frac{d}{dx}(2x^2 - 2) = 4x \).

Step 5: Apply the chain rule

Substitute the derivatives from Steps 2, 3, and 4 into the chain rule:\[\frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \cdot 2u = 2 \sec(v(x)) \cdot \sec(v(x)) \tan(v(x)) \cdot 4x. \] Simplifying gives:\[\frac{df}{dx} = 8x \sec^2(v(x)) \tan(v(x)).\]

Step 6: Substitute back v(x)

Substitute \( v(x) = 2x^2 - 2 \) back into the expression: \[\frac{df}{dx} = 8x \sec^2(2x^2 - 2) \tan(2x^2 - 2).\]

Key Concepts

Understanding the Chain Rule in Calculus

The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function is a function made up of two or more other functions. In simpler terms, it is a function inside another function.
Consider a function defined as \[ h(x) = f(g(x)) \] where one function, \( g(x) \), is the "inner function" and \( f \) is the "outer function."
To find the derivative of \( h(x) \), the chain rule tells us to differentiate the outer function with respect to the inner function, and multiply it by the derivative of the inner function itself: \[ \frac{dh}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \] ##### How the Chain Rule Simplifies Differentiation- *Identify Functions*: Recognize the outer and inner functions.
- *Differentiate Each Part*: First, differentiate the outer function and then the inner one.
- *Combine with Chain Rule*: Multiply the derived parts as instructed by the chain rule.
Applying the chain rule makes complex derivatives easy to handle by breaking down the process into manageable steps.

Grasping the Concept of Derivatives

A derivative represents the rate of change of a function with respect to a variable. If we think of a function as a curve drawn on a graph, the derivative tells us the slope of the curve at any given point.
Mathematically, if \( y = f(x) \), the derivative is usually denoted as \( \frac{dy}{dx} \).
##### Why Derivatives Matter- *Identify Slope*: Derivatives help find the slope of a function at a particular point.
- *Predict Behavior*: Understanding derivatives enables prediction of a function's behavior and how it changes.
##### Basic Derivatives You Should Know- *Constant Function*: The derivative of any constant is zero.- *Power Rule*: For any power function \( x^n \), the derivative is \( nx^{n-1} \).
- *Exponential and Logarithmic Functions*: These have their own specific rules for differentiation.
Derivatives are crucial in finding how functions respond to changes in their variables, making them useful in various fields such as physics and engineering.

Exploring Trigonometric Functions in Calculus

Trigonometric functions like sine, cosine, and tangent are fundamental in calculus, especially when dealing with periodic functions that express wave-like patterns.
Each trigonometric function has its own derivative, which helps us understand how these functions change with respect to an angle.##### Common Trigonometric Derivatives:- *Sine* \( (\sin x) \): Derivative is \( \cos x \)
- *Cosine* \( (\cos x) \): Derivative is \( -\sin x \)
- *Tangent* \( (\tan x) \): Derivative is \( \sec^2 x \)
Understanding these derivatives assists in solving complex calculus problems where trig functions are involved.
In composite functions or more advanced math problems, knowing these derivatives is essential.##### Applications of Trigonometric Derivatives- *Wave Equations*: Describe physical waves and vibrations.
- *Periodic Functions*: Adjust models for repetitive behaviors such as pendulum motion.

Grasping the derivatives of trigonometric functions is vital in calculus as it allows accurate modeling and solving of geometric and physical problems.