Problem 16 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln (\sin x)$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = \cot x\).

Step 1: Identify the Function

We are given the function \(y = \ln(\sin x)\). Our task is to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

Step 2: Apply the Chain Rule

To differentiate \(y = \ln(\sin x)\), we identify it as a composition of functions: \(u = \sin x\) and \(y = \ln(u)\). The chain rule states that \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).

Step 3: Differentiate the Outer Function

First, differentiate the outer function \(y = \ln(u)\) with respect to \(u\). This gives us \(\frac{dy}{du} = \frac{1}{u}\).

Step 4: Differentiate the Inner Function

Next, differentiate the inner function \(u = \sin x\) with respect to \(x\). We find \(\frac{du}{dx} = \cos x\).

Step 5: Combine the Results

Now apply the chain rule by multiplying the two derivatives obtained: \(\frac{dy}{dx} = \frac{1}{\sin x} \times \cos x = \frac{\cos x}{\sin x}\).

Step 6: Simplify the Expression

We simplify \(\frac{\cos x}{\sin x}\) using the trigonometric identity for cotangent: \(\frac{dy}{dx} = \cot x\).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. It comes into play when you have a function inside another function. In our exercise, the function is given as \( y = \ln(\sin x) \). Here, we recognize it as a composition of two functions:

• the outer function is \( \ln(u) \)
• the inner function is \( u = \sin x \)

The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of the outer function \( \frac{dy}{du} \) by the derivative of the inner function \( \frac{du}{dx} \). So, for our function:\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]This breaks the problem into manageable chunks, making differentiation of complex functions faster and easier.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) frequently appear in calculus, especially in differentiation problems. When dealing with trigonometric functions, it's crucial to know their derivatives. For instance:

• The derivative of \( \sin x \) is \( \cos x \)
• The derivative of \( \cos x \) is \(-\sin x \)

In our problem, the inner function \( u = \sin x \) was differentiated to obtain \( \frac{du}{dx} = \cos x \). This result is essential when applying the chain rule, as it gives us part of what we need to find the complete derivative \( \frac{dy}{dx} \). Understanding these basic trigonometric derivatives is key to tackling more complicated calculus problems involving trigonometric functions.

Logarithmic Differentiation

Logarithmic differentiation is a method often used when dealing with functions that involve logarithms, like \( \ln(x) \). In the exercise, it helps simplify the differentiation of \( y = \ln(\sin x) \) due to the logarithmic nature of the outer function:

• When you differentiate \( \ln(u) \) with respect to \( u \), the result is \( \frac{1}{u} \).

Thus, when we apply this to our function, the outer derivative becomes \( \frac{dy}{du} = \frac{1}{\sin x} \). By combining this result with the inner derivative derived from trigonometric differentiation, we get the full derivative:\[\frac{dy}{dx} = \frac{1}{\sin x} \times \cos x\]This simplifies due to trigonometric identities to \( \cot x \), providing us with a neat solution. Logarithmic differentiation is particularly handy in breaking down seemingly complex expressions into simpler parts.