Problem 22 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=5^{-\cos 2 t}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = 2 \ln(5) \sin(2t) 5^{-\cos(2t)} \).

Step 1: Recognize the Function Type

The function given is an exponential function where the base is a constant (5) and the exponent is a trigonometric function (-cos 2t). The standard form of an exponential function is \( a^{b(x)} \), where \( a \) is a constant base, and \( b(x) \) is an exponent depending on x.

Step 2: Apply Logarithmic Differentiation

To differentiate this function, use the formula: \( \frac{d}{dx}a^{b(x)} = a^{b(x)} \ln(a) \cdot \frac{d}{dx}[b(x)] \). Here, \( a=5 \) and \( b(x) = -\cos(2t) \).

Step 3: Differentiate the Exponent

Differentiate the exponent \(-\cos(2t)\) with respect to \(t\). Using the chain rule, \( \frac{d}{dt}[-\cos(2t)] = -[-\sin(2t)] \cdot 2 \). This simplifies to \(2\sin(2t)\).

Step 4: Substitute Back into the Derivative Formula

Substitute the calculations from Step 3 back into the derivative formula. The derivative is:\[ \frac{dy}{dt} = 5^{-\cos(2t)} \ln(5) \cdot 2\sin(2t) \]

Step 5: Simplify the Expression

Finally, simplify the expression:\[ \frac{dy}{dt} = 2 \cdot \ln(5) \cdot \sin(2t) \cdot 5^{-\cos(2t)} \]. This is the derivative of \( y \) with respect to \( t \).

Key Concepts

Exponential Function

Exponential functions are a critical concept in calculus and appear in many mathematical contexts. An exponential function is usually expressed in the form \( a^{b(x)} \), where \( a \) is a constant base greater than zero, and \( b(x) \) is any real function dependent on \( x \).
What makes exponential functions stand out is their unique property: they grow or decay at rates proportional to their value. This results in fascinating mathematical behavior, essential in fields like finance for compound interest calculations, biology for population dynamics, and physics for radioactive decay.
In our exercise, \( y = 5^{- ext{cos}(2t)} \), "5" is the constant base and "\(-\cos(2t)\)" is a trigonometric function in the exponent. This combination complicates differentiation using standard rules.
To find the derivative of such functions, log differentiation, a powerful tool, simplifies the process significantly, especially when composite functions are involved.

Chain Rule

The chain rule is an essential differentiation technique when dealing with composite functions. It allows us to differentiate functions nested within one another. The chain rule states: if you have a function \( f(g(x)) \), then the derivative is \( f'(g(x)) \times g'(x) \).
In the exercise, the exponent of our function \( y = 5^{- ext{cos}(2t)} \) is the composite function \(- ext{cos}(2t)\). To differentiate this, we need the chain rule because it consists of a cosine function of \(2t\).
Here’s how it applies: first, differentiate \(- ext{cos}(2t)\), giving \(2\sin(2t)\). This step is crucial parallel to applying the chain rule, which breaks down the differentiation process into manageable steps.
Understanding the chain rule unlocks the ability to tackle various complex functions by breaking them down into simpler parts.

Trigonometric Differentiation

Trigonometric differentiation involves finding the derivatives of trigonometric functions such as sine, cosine, and tangent. Trigonometric functions appear frequently in calculus, particularly in problems involving oscillatory behavior, such as waves.
In our problem, we need to differentiate \(-\text{cos}(2t)\), which is our exponent in the exponential function. The derivative of \(\cos(2t)\) is \(-\sin(2t)\), but due to the chain rule and constant multiplication, the derivative becomes \(2\sin(2t)\).
Key derivatives to remember in trigonometric differentiation include:

• \( \frac{d}{dx}[\sin(x)] = \cos(x) \)
• \( \frac{d}{dx}[\cos(x)] = -\sin(x) \)
• \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \)

These rules and proper application of the chain rule, as seen in the exercise, help evaluate the derivatives of trigonometric expressions effectively in various contexts.