Question

Calculate the derivative of the following functions. $$y=\cos x \ln \left(\cos ^{2} x\right)$$

Short answer

Answer: The derivative of the given function is \(y' = -\sin x\ln (\cos^2 x) -2\sin x\cos x\).

Step-by-step solution

Identify the Functions

We have a product of two functions, \(f(x) = \cos x\) and \(g(x) = \ln (\cos^2 x)\).

Apply the Product Rule

The product rule states that if we have two functions, \(u(x)\) and \(v(x)\), then the derivative of their product is given by: \((uv)' = u'v + uv'\). We will apply this rule to our functions \(f(x)\) and \(g(x)\). But before that, we need to find the derivatives of these functions.

Differentiate f(x)

We have \(f(x) = \cos x\). The derivative of the cosine function is \(-\sin x\). So, \(f'(x) = -\sin x\).

Differentiate g(x)

We have \(g(x) = \ln (\cos^2 x)\). To differentiate this function, we'll use the chain rule. The chain rule states that if we have a function \(h(g(x))\), then the derivative of this function is given by: \(\frac{d}{dx}h(g(x)) = h'(g(x)) \cdot g'(x)\). So, we have the outer function \(h(u) = \ln u\) and the inner function \(u(x) = \cos^2 x\). Differentiate the outer function with respect to u: \(\frac{dh}{du} = \frac{1}{u}\). Now, differentiate the inner function with respect to x: \(\frac{du}{dx} = \frac{d}{dx}(\cos^2 x) = 2\cos x \cdot (-\sin x) = -2\cos x\sin x\). Applying the chain rule, we get: $$g'(x) = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{1}{\cos^2 x} \cdot (-2 \cos x \sin x) = -2\sin x\sec^2 x$$.

Apply Product Rule to find y'

Now, we apply the product rule to find the derivative of y: $$y' = f'(x)g(x) + f(x)g'(x) = (-\sin x)\ln (\cos^2 x) + (\cos x)(-2\sin x\sec^2 x)$$. Simplify the expression: $$y' = -\sin x\ln (\cos^2 x) -2\sin x\cos x$$. The derivative of the function \(y = \cos x \ln(\cos^2 x)\) is $$y' = -\sin x\ln (\cos^2 x) -2\sin x\cos x$$.

Key concepts

Product Rule

The product rule is essential when dealing with the derivatives of products of functions. When you have two functions, say \( u(x) \) and \( v(x) \), their product \( y = u(x) \cdot v(x) \) requires the product rule for differentiation. The rule states that the derivative of the product \( (uv)' \) is given by:

• \( u'(x) \times v(x) \) + \( u(x) \times v'(x) \)

This means you must find the derivative of each function individually and then combine them as described above. It's useful in problems where functions are multiplied together, such as trigonometric functions multiplied with logarithmic functions. By treating each part separately, you break down the complexity of derivatives into manageable parts. For example, using the product rule helps us find the derivative of \( y = \cos x \ln(\cos^2 x) \) by letting \( f(x) = \cos x \) and \( g(x) = \ln(\cos^2 x) \), and then applying the rule.

Chain Rule

The chain rule is a powerful tool for differentiating composite functions. A composite function is when you have one function nested inside another, like \( y = h(g(x)) \). The chain rule helps you differentiate such functions by considering both the inner and outer functions. The rule is described as follows:

• First, differentiate the outer function with respect to its argument.
• Then, differentiate the inner function with respect to \( x \).
• Finally, multiply these derivatives together.

For instance, in the expression \( g(x) = \ln(\cos^2 x) \), the outer function \( h(u) = \ln u \) and the inner function \( u(x) = \cos^2 x \). Initially, differentiate \( \ln u \) to get \( \frac{1}{u} \), and differentiate \( \cos^2 x \) using the power and chain rules, resulting in \( -2\cos x\sin x \). Combining these using the chain rule, you derive \( g'(x) = -2\sin x \sec^2 x \). The chain rule efficiently handles the complexities of multiple function layers, as seen in our example.

Trigonometric Functions

Understanding trigonometric functions, such as sine and cosine, is crucial for calculus, particularly in derivatives. These functions describe periodic patterns, making them pivotal in physics and engineering. In the derivative context, remember:

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

In problems involving products or compositions of such functions, recognizing these basic derivatives is the first step. For instance, when dealing with \( f(x) = \cos x \), its derivative \( f'(x) = -\sin x \) becomes part of using the product rule when differentiating \( y = \cos x\ln(\cos^2 x) \). Moreover, when deriving derivatives of composite trigonometric functions, tools like the chain rule often come into play. As seen in \( g(x) = \ln(\cos^2 x) \), understanding the behavior of \( \cos x \) helps simplify and solve differentiation, demonstrating the trigonometric functions' indispensable role in calculus.