Problem 55 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative is \(-2\tan(\theta)\).

Step 1: Understand the Inner Function

First, identify the function inside the natural logarithm. Here, the function is \(g(\theta) = \cos^{2}(\theta)\). This means that \(y = \ln(g(\theta))\).

Step 2: Use the Chain Rule

Since the function \(y\) is composed of \(\ln(u)\) where \(u = g(\theta) = \cos^{2}(\theta)\), apply the chain rule \(\frac{dy}{d\theta} = \frac{d}{du}(\ln u) \cdot \frac{du}{d\theta}\).

Step 3: Differentiate the Outer Function

The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). Therefore, \(\frac{d}{du}(\ln u) = \frac{1}{\cos^{2}(\theta)}\).

Step 4: Differentiate the Inner Function

Differentiate \(\cos^{2}(\theta)\) with respect to \(\theta\). Use the chain rule again, noting that \(\cos^{2}(\theta) = (\cos(\theta))^2\), so \(\frac{du}{d\theta} = 2\cos(\theta)(-\sin(\theta))=-2\cos(\theta)\sin(\theta)\).

Step 5: Combine Results Using the Chain Rule

Following the chain rule, multiply the derivatives from Steps 3 and 4: \(\frac{dy}{d\theta} = \frac{1}{\cos^{2}(\theta)} \times (-2\cos(\theta)\sin(\theta))\).

Step 6: Simplify the Expression

Simplify \(\frac{dy}{d\theta} = \frac{-2\cos(\theta)\sin(\theta)}{\cos^{2}(\theta)}\). Simplifying further, this becomes \(\frac{-2\sin(\theta)}{\cos(\theta)}\) which is \(-2\tan(\theta)\) because \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a mathematical function that serves as the inverse of the exponential function \(e^x\). It is used extensively in calculus and mathematical modeling, particularly with scenarios involving growth and decay.

• The natural logarithm grows very slowly compared to polynomial functions.
• It is only defined for positive input values, i.e., \(x>0\). This is because you can't take the logarithm of zero or a negative number in the real number system.
• The derivative of the natural logarithm is crucial in calculus: \(\frac{d}{dx}(\ln(x)) = \frac{1}{x}\).

In this exercise, we deal with \(\ln(\cos^2(\theta))\). This involves applying the natural logarithm to an expression derived from trigonometric functions. It is key to understand that the derivative formula for \(\ln(x)\) helps guide our differentiation steps.

Chain Rule

The chain rule is a fundamental tool in calculus used to find derivatives of composite functions. Whenever a function is nested inside another function, the chain rule is used to differentiate it.

• The rule states: if \(y = f(g(x))\), then the derivative \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).
• Each part of the composite function needs to be understood separately. Start by finding the derivative of the outer function, then multiply by the derivative of the inner function.

For the given problem, the outer function is the natural logarithm \(\ln(u)\) and the inner function is \(\cos^2(\theta)\). Applying the chain rule involves first differentiating \(\ln(u)\) with respect to \(u\), and then multiplying by the derivative of \(u = \cos^2(\theta)\) with respect to \(\theta\). This step-by-step approach helps break down complex differentiation tasks into manageable pieces.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent, play an essential role in calculus. They are periodic functions commonly encountered in problems involving oscillatory behavior, right triangles, and unit circles.

• The basic derivatives to remember are: \(\frac{d}{d\theta}(\sin(\theta)) = \cos(\theta)\) and \(\frac{d}{d\theta}(\cos(\theta)) = -\sin(\theta)\).
• Combining these functions often requires additional calculus rules, such as the chain rule, or product rule, for derivatives.

In the exercise, \(\cos^2(\theta)\) is the square of the cosine function, which needs the application of the chain rule to differentiate correctly. By understanding that \(\cos^2(\theta) = (\cos(\theta))^2\), we can find its derivative as \(-2\cos(\theta)\sin(\theta)\). This is due to applying the derivative chain on the cosine squared function. The final result involving \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\) showcases the transformation from a trigonometric expression into another familiar form.