Question

Apply the Chain Rule more than once to find the indicated derivative. \(\frac{d}{d t}\left\\{\cos ^{2}[\cos (\cos t)]\right\\}\)

Short answer

\( 2 \sin(t) \sin(\cos(t)) \cdot \cos(\cos(\cos(t))) \sin(\cos(\cos(t))) \)

Step-by-step solution

Identify the Outer Functions

The expression can be broken down into functions: the outermost function is \((t) = ( ext{cos}(x))^2\), where \(x = ext{cos}( ext{cos}(t))\). The second function is \(g(x) = ext{cos}(x)\), where \(x = ext{cos}(t)\). The innermost function is \(h(t) = ext{cos}(t)\).

Apply the Chain Rule (Outer Function)

Differentiate the outer function \(f(x) = ( ext{cos}(x))^2\) with respect to \(x\): \(f'(x) = 2 ext{cos}(x)(- ext{sin}(x))\).

Apply the Chain Rule (Second Layer)

Differentiate the middle function \(g(x) = ext{cos}(x)\) with respect to \(x\): \(g'(x) = - ext{sin}(x)\).

Apply the Chain Rule (Innermost Function)

Differentiate the innermost function \(h(t) = ext{cos}(t)\) with respect to \(t\): \(h'(t) = - ext{sin}(t)\).

Compose Derivatives Together

Stack the derivatives together by multiplying them: \[\frac{d}{dt} = h'(t) imes g'(t) imes f'(g(h(t)))\]Substituting in the derivatives: \[= [- ext{sin}(t)] imes [- ext{sin}( ext{cos}(t))] imes [2 ext{cos}( ext{cos}( ext{cos}(t)))(- ext{sin}( ext{cos}( ext{cos}(t))))]\]

Simplify the Expression

Simplify the final expression of the derivative: \[\frac{d}{dt}ig(\cos^{2}[\cos(\cos t)]\big) = 2 \text{sin}(t) \cdot \text{sin}(\cos(t)) \cdot \cos(\cos(\cos(t))) \cdot \text{sin}(\cos(\cos(t)))\]

Key concepts

calculus

Calculus forms the foundation of understanding changes in mathematics and real-world phenomena. At its core, calculus explores how things change and provides tools to describe those changes. It encompasses two significant branches: differential calculus and integral calculus. Differential calculus focuses on finding how functions change, which involves finding their derivatives, or rates of change. This is crucial when analyzing anything that evolves over time or in response to other factors, such as speed, growth, and more.

In calculus, derivatives are essential tools that tell us the rate at which one quantity changes with respect to another. They are like mathematical zoom lenses that provide detailed insights into complex relationships. These powerful techniques allow us to solve real-world problems involving motion, optimization, and even predicting future trends.

If you're beginning with calculus, remember it's about understanding the language of change. It's about breaking down complicated motions into more manageable pieces to see the relationships between different variables. This sets the stage for more advanced topics and applications in mathematics and sciences.

derivative techniques

Derivative techniques are ways to find the rate of change of functions, and one of the most versatile of these techniques is the Chain Rule. The Chain Rule allows us to differentiate compositions of functions, which are expressions where one function is nested inside another. This method aligns with the idea of peeling back layers of complex expressions, which makes it easier to handle them mathematically.

When you encounter functions wrapped inside each other, like in our exercise above, the Chain Rule helps you tackle them layer by layer:

• Identify the layers: Break down the function into its outer and inner layers.
• Differentiate each layer: Start from the outermost function and work inward, applying the required differentiation rules.
• Multiply the derivatives: Combine the individual derivatives by multiplying them in sequence, starting from the innermost layer to the outermost.

The Chain Rule becomes especially effective with complicated functions involving trigonometric expressions, exponents, and more. Mastering this rule opens up a wide array of functions that you can now differentiate, propelling you further in your mathematical journey. Practice breaking down functions and applying the Chain Rule to build confidence and proficiency.

trigonometric functions

Trigonometric functions are mathematical functions of an angle, and they play a vital role in calculus due to their periodic nature. Commonly known as sine, cosine, and tangent, these functions arise naturally in a variety of sciences and are familiar from the study of triangles and circles.

In calculus, the derivatives of trigonometric functions are crucial and help in tackling tasks involving oscillating motions or waves, such as sound waves or light patterns.

• Sine (\( ext{sin}(x)\)): Differentiates to cosine, \( ext{cos}(x)\).
• Cosine (\( ext{cos}(x)\)): Differentiates to negative sine, \(- ext{sin}(x)\).

In our given exercise, these derivatives were used at multiple layers. Recognizing how each layer functions with respect to its containing functions is essential. By using trigonometric derivatives, the Chain Rule aids in simplifying and finding the overall rate of change.

Understanding these basics can help you speed through problems involving trigonometric functions. As you progress, you'll see that these building blocks will be used over and over in many different contexts. Keep in mind their periodic properties and how they interact in composite functions, like in the exercise at hand.