Problem 58 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(2 e^{-t} \sin t\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \cot t - 1 \)

Step 1: Identify the Derivative Formula

To find the derivative of a logarithmic function, we can use the derivative formula for the natural log function: \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \), where \( u \) is a function of \( x \).

Step 2: Identify \( u \)

In the given function \( y = \ln(2e^{-t} \sin t) \), identify \( u \) as \( u = 2e^{-t} \sin t \).

Step 3: Differentiate \( u \)

Find the derivative of \( u \): \( u = 2e^{-t} \sin t \). Use the product rule \( \frac{d}{dt}(fg) = f'g + fg' \) where \( f = 2e^{-t} \) and \( g = \sin t \).

Step 4: Apply the Product Rule

Calculate \( f' \) and \( g' \):- \( f = 2e^{-t} \), so \( f' = -2e^{-t} \) (derivative of an exponential function: \( \frac{d}{dt} e^{-t} = -e^{-t} \)).- \( g = \sin t \), so \( g' = \cos t \) (derivative of sine is cosine).Thus, \( \frac{du}{dt} = (-2e^{-t}) \sin t + (2e^{-t}) \cos t \).

Step 5: Simplify \( \frac{du}{dt} \)

Factor out \( 2e^{-t} \):\[ \frac{du}{dt} = 2e^{-t}(-\sin t + \cos t) \].

Step 6: Apply the Chain Rule to \( y \)

Use the formula from Step 1:\[ \frac{dy}{dt} = \frac{1}{u} \cdot \frac{du}{dt} = \frac{1}{2e^{-t} \sin t} \cdot 2e^{-t}(-\sin t + \cos t) \].

Step 7: Simplify the Expression

Cancel out \( 2e^{-t} \) to obtain:\[ \frac{dy}{dt} = \frac{-\sin t + \cos t}{\sin t} \].

Step 8: Final Simplification

Rearrange the expression:\[ \frac{dy}{dt} = \frac{\cos t}{\sin t} - 1 = \cot t - 1 \].

Key Concepts

Product Rule in Calculus

The Product Rule in calculus is a fundamental differentiation rule that arises when you need to find the derivative of the product of two functions. This rule is essential because many functions we encounter are products of simpler functions. Think about it like a recipe where we need both ingredients to make a delicious dish!When you have two functions, say \( f(t) \) and \( g(t) \), and you need to find the derivative of their product \( f(t)g(t) \), you use the Product Rule formula:

• \( \frac{d}{dt}(f(t)g(t)) = f'(t)g(t) + f(t)g'(t) \)

This means we differentiate each function separately and then combine them according to the rule.In the given problem, we have a product \( u = 2e^{-t} \sin t \). Here,

• \( f(t) = 2e^{-t} \) and \( f'(t) = -2e^{-t} \)
• \( g(t) = \sin t \) and \( g'(t) = \cos t \)

Using the Product Rule, the derivative \( \frac{du}{dt} \) becomes:\[ -2e^{-t} \sin t + 2e^{-t} \cos t \] This skill is very advantageous when working with complex logarithmic functions, as it breaks the problem into manageable parts.

Chain Rule in Calculus

The Chain Rule is another indispensable tool in calculus, primarily used for differentiating composite functions. A composite function is essentially a function within another function, kind of like those Russian nesting dolls where each layer reveals more!The Chain Rule allows us to find the derivative of a composition by taking the derivative of the outer function and multiplying it by the derivative of the inner function. The formula can be understood as:

• \( \frac{d}{dt}[h(g(t))] = h'(g(t)) \cdot g'(t) \)

In our exercise, to find \( \frac{dy}{dt} \), we first need the derivative of the outer function \( \ln(u) \), and then multiply by the derivative of \( u \) itself (which was obtained using the Product Rule). Therefore:\[ \frac{dy}{dt} = \frac{1}{u} \cdot \frac{du}{dt} \]This is exactly where the Chain Rule shines — it's like the glue holding the rest of the problem together, ensuring that both the outer and inner functions are considered when finding the derivative.

Simplification of Trigonometric Expressions

Simplification is the art of making a mathematical expression more compact and easier to understand. It’s like tidying up a room and making everything neat. With trigonometric expressions, this process can involve using identities or algebraic manipulation.After applying the Chain Rule, the expression we arrived at was:\[ \frac{dy}{dt} = \frac{-\sin t + \cos t}{\sin t} \]To simplify this, we can break it down further by dividing each term by \( \sin t \):

• \( \frac{-\sin t}{\sin t} = -1 \)
• \( \frac{\cos t}{\sin t} = \cot t \)

This transforms the derivative into the more elegant form:\[ \frac{dy}{dt} = \cot t - 1 \]This final expression uses the cotangent in place of the fraction, making it not only simpler but also more interpretable for further analysis. Simplification lets you focus on understanding rather than wading through complex-looking math, ensuring clarity in your results.