Problem 66 - Thomas Calculus in SI Units

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

Step 1: Identify the Function Structure

The function given is \( y = 5^{- ext{cos}(2t)} \). This is an exponential function, where the base is constant (5) and the exponent is a trigonometric function of \( t \). Our aim is to find \( \frac{dy}{dt} \).

Step 2: Apply the Exponential Function Rule

The derivative of \( a^{u} \) with respect to \( x \), where \( a \) is a constant and \( u \) is a differentiable function of \( x \), is \( a^{u} \ln{a} \cdot \frac{du}{dx} \). Here, \( a = 5 \) and \( u = -\cos(2t) \).

Step 3: Differentiate the Exponent

Find \( \frac{d(-\cos(2t))}{dt} \). First, apply the chain rule: \( \frac{d(-\cos(2t))}{dt} = -(\frac{d(\cos(2t))}{d(2t)}) \cdot \frac{d(2t)}{dt} \). The derivative of \( \cos(2t) \) with respect to \( 2t \) is \( -\sin(2t) \), and the derivative of \( 2t \) is 2. Thus, \( \frac{d(-\cos(2t))}{dt} = 2\sin(2t) \).

Step 4: Substitute Back to Find the Derivative

Using the formula for the derivative of \( a^u \), substitute the values: \( \frac{dy}{dt} = 5^{- ext{cos}(2t)} \ln(5) \cdot 2\sin(2t) \).

Step 5: Simplify the Expression

Combine the terms to get the final expression for the derivative: \( \frac{dy}{dt} = 2\ln(5) \cdot 5^{- ext{cos}(2t)} \cdot \sin(2t) \). This is the simplified form of the derivative of the function.

Key Concepts

Chain Rule in Differentiation

The chain rule is a fundamental tool in calculus used to differentiate composite functions. A composite function is essentially a function within another function, and the chain rule helps us deal with these layers effectively. If we have two functions, let's say \( f(x) \) and \( g(x) \), then the composite function is \( f(g(x)) \). The chain rule states that the derivative of \( f(g(x)) \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \).

In our specific problem, we are differentiating an exponential function where the exponent itself is a trigonometric function. For the expression \( y = 5^{-\cos(2t)} \), the inner function here is \( -\cos(2t) \). When you differentiate a function like this, you need to apply the chain rule to the exponent part. Instead of directly using exponential rules, we must first consider the differentiation of the inner part, \( -\cos(2t) \), which leads us to our next concept.

Trigonometric Functions

Trigonometric functions are essential in calculus, especially when dealing with derivatives involving sine, cosine, and other related functions. These are periodic functions that model waves and oscillations. The basic derivatives you should remember are:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In the exercise at hand, we handle \( \cos(2t) \). Differentiating this requires special attention due to the presence of \( 2t \) inside the function. First, differentiate the \( \cos(2t) \) to get \( -\sin(2t) \). Next, apply the derivative of the inner function \( 2t \), which is simply 2. Multiplying these results gives \( 2\sin(2t) \) for the derivative of \( -\cos(2t) \). This result was crucial in applying the chain rule correctly in the given function.

Exponential Functions

Exponential functions have a constant as a base and a variable exponent. The most straightforward representation is \( a^x \), where \( a \) is the base and \( x \) is the exponent. With the exponential function, the rule to differentiate \( a^u \), where \( u \) is a function of \( x \), involves multiplying by the natural logarithm of the base and the derivative of the exponent: \( \frac{d}{dx}(a^u) = a^u \ln(a) \cdot \frac{du}{dx} \). In your exercise, the base number is 5, and the exponent is \(-\cos(2t)\), a trigonometric function. To handle this, calculate \( 5^{-\cos(2t)} \ln(5) \cdot \frac{d(-\cos(2t))}{dt} \). This beautifully integrates both the chain rule and knowledge of trigonometric derivatives.

Ultimately, using this process gives you \( 2\ln(5) \cdot 5^{-\cos(2t)} \cdot \sin(2t) \), which is the core insight that ties together the exponential nature, the logarithmic relationship, and the tailored differentiation method for the exponent.