Problem 70 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the function form

The function given is \( y = 5^{- ext{cos} \, 2t} \). It is an exponential function with base 5 and exponent \(-\text{cos} \, 2t\). We need to find its derivative with respect to \( t \).

Step 2: Apply the logarithmic differentiation

To differentiate \( y = 5^{-\text{cos} \, 2t} \), we take the natural logarithm on both sides: \( \ln y = \ln(5^{-\text{cos} \, 2t}) \). This simplifies to \( \ln y = -\text{cos} \, 2t \cdot \ln 5 \).

Step 3: Differentiate both sides with respect to \( t \)

Using implicit differentiation, we find the derivative of \( \ln y \) with respect to \( t \): \( \frac{1}{y} \frac{dy}{dt} = -\ln 5 \cdot \frac{d}{dt}(\text{cos} \, 2t) \).

Step 4: Calculate the derivative of \( \text{cos} \, 2t \)

The derivative of \( \text{cos} \, 2t \) with respect to \( t \) is \( -2 \sin 2t \) using the chain rule.

Step 5: Substitute back and solve for \( \frac{dy}{dt} \)

Substitute \( \frac{d}{dt}(\text{cos} \, 2t) = -2 \sin 2t \) into the equation: \( \frac{1}{y} \frac{dy}{dt} = 2 \ln 5 \sin 2t \). Thus, \( \frac{dy}{dt} = y \cdot 2 \ln 5 \sin 2t \).

Step 6: Substitute the original function back into the derivative

Replace \( y \) with \( 5^{-\text{cos} \, 2t} \) to express \( \frac{dy}{dt} \) in terms of \( t \): \( \frac{dy}{dt} = 5^{-\text{cos} \, 2t} \cdot 2 \ln 5 \sin 2t \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used for finding derivatives of functions that are difficult to differentiate using traditional methods. It's especially useful when dealing with products, quotients, or powers of functions. By applying the natural logarithm to both sides of an equation, you simplify the process of differentiating complex expressions.

In the given exercise, we have the function \( y = 5^{-\cos 2t} \). This function's exponent involves a trigonometric function, making direct differentiation tricky. To handle this, we take the natural logarithm of both sides:

• Apply the logarithm: \( \ln y = \ln(5^{-\cos 2t}) \)
• Use logarithm properties: \( \ln y = -\cos 2t \cdot \ln 5 \)

Once the function is in logarithmic form, we differentiate implicitly with respect to \( t \). This yields the derivative of the left side—\( \ln y \)—in terms of \( y \), and simplifies the right side's differentiation. Logarithmic differentiation thus allows us to manage the complexities of the exponential function more effectively.

Exponential Function

Exponential functions can be tricky to differentiate, particularly when the exponent contains a variable. Here, our function \( y = 5^{-\cos 2t} \) is an example of an exponential function where both the base and the exponent involve complexities.

In this exercise, the base is the constant 5, while the exponent \(-\cos 2t\) incorporates a trigonometric element that varies with \( t \). When differentiating exponential functions, recognizing and isolating the constant base from the variable exponent is crucial. Once identified, applying the natural logarithm to transform the function into a more manageable form becomes possible:

• The constant base remains as it is in logarithmic differentiation.
• The exponent \(-\cos 2t\), with its variable nature, becomes the key focus for differentiation.

Breaking it down in this manner facilitates differentiating complex exponential functions by treating them in parts—turning them from intimidating powerhouses into manageable entities.

Chain Rule

The chain rule is an essential tool in calculus for finding derivatives of composite functions. It allows us to handle functions within functions, like \( -\cos 2t \) that appears in the exponent of our original function. To apply the chain rule effectively, break down each function in the composition and proceed step-by-step.

Here's how it is used in our exercise:

• The outer function: Exponential form \(5^{u}\) where \(u = -\cos 2t\).
• The inner function: \(-\cos 2t\), which is a trigonometric function whose derivative is necessary.

Differentiating the outer function first gives us the form involving \(u\), while applying the chain rule to the inner function involves finding the derivative of \(-\cos 2t\). Applying this rule yields:

• Derivative of \(\cos 2t\) is \(-2\sin 2t\), using the derivative rule for cosine and the multiplication chain rule for 2.

By combining these derivatives, we obtain the full derivative of the composite function \( y = 5^{-\cos 2t} \). The chain rule shines in its ability to unravel the complexities of nested functions, making differentiation straightforward and systematic.