Differentiating \(f(x) = \cos(\tan 3x) + \sin(\tan 3x)\)

To find the derivative of the given function \(f(x) = \cos(\tan 3x) + \sin(\tan 3x)\), we need to apply the rules of differentiation, specifically the chain rule, to each term separately. The differentiation process will involve breaking down each composite function into its inner and outer parts.

Understanding the Chain Rule for Differentiation

The chain rule is a fundamental rule in calculus used for differentiating composite functions. A composite function is essentially a function inside another function. If we have a function \(y = g(u)\) where \(u = h(x)\), then the derivative of \(y\) with respect to \(x\) is given by the formula:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

In our problem, \(f(x)\) consists of two main terms, each requiring the application of the chain rule:

• The first term is \(\cos(\tan 3x)\).
• The second term is \(\sin(\tan 3x)\).

We will differentiate each term step-by-step and then sum their derivatives to find the derivative of \(f(x)\).

Differentiating the First Term: \(\cos(\tan 3x)\)

Let's find the derivative of the first term, \(y_1 = \cos(\tan 3x)\). We identify the layers of functions here:

• The outermost function is \(\cos(u)\), where \(u\) represents \(\tan 3x\).
• The next inner function is \(\tan(v)\), where \(v\) represents \(3x\).
• The innermost function is \(3x\).

Applying the chain rule:

1. Differentiate the outermost function: The derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\). Substituting \(u = \tan 3x\) back, we get \(-\sin(\tan 3x)\).
2. Differentiate the middle function: Next, we need the derivative of \(\tan 3x\) with respect to \(x\). This also requires the chain rule.
   • The derivative of \(\tan(v)\) with respect to \(v\) is \(\sec^2(v)\). Substituting \(v = 3x\) back, we get \(\sec^2(3x)\).
   • The derivative of the innermost function \(3x\) with respect to \(x\) is \(3\).
   So, the derivative of \(\tan 3x\) is \(\sec^2(3x) \cdot 3 = 3\sec^2(3x)\).

Now, multiply the derivatives of the layers to get the derivative of \(\cos(\tan 3x)\):

\[ \frac{d}{dx}(\cos(\tan 3x)) = -\sin(\tan 3x) \cdot 3\sec^2(3x) = -3\sec^2(3x)\sin(\tan 3x) \]

Differentiating the Second Term: \(\sin(\tan 3x)\)

Next, let's find the derivative of the second term, \(y_2 = \sin(\tan 3x)\). Similar to the first term, we identify the layers:

• The outermost function is \(\sin(u)\), where \(u\) represents \(\tan 3x\).
• The next inner function is \(\tan(v)\), where \(v\) represents \(3x\).
• The innermost function is \(3x\).

Applying the chain rule:

1. Differentiate the outermost function: The derivative of \(\sin(u)\) with respect to \(u\) is \(\cos(u)\). Substituting \(u = \tan 3x\) back, we get \(\cos(\tan 3x)\).
2. Differentiate the middle function: As calculated for the first term, the derivative of \(\tan 3x\) with respect to \(x\) is \(3\sec^2(3x)\).

Now, multiply these derivatives to find the derivative of \(\sin(\tan 3x)\):

\[ \frac{d}{dx}(\sin(\tan 3x)) = \cos(\tan 3x) \cdot 3\sec^2(3x) = 3\sec^2(3x)\cos(\tan 3x) \]

Combining the Derivatives for \(f(x)\)

Since \(f(x) = \cos(\tan 3x) + \sin(\tan 3x)\), its total derivative \(f'(x)\) is the sum of the derivatives of its individual terms:

\[ f'(x) = \frac{d}{dx}(\cos(\tan 3x)) + \frac{d}{dx}(\sin(\tan 3x)) \]

Substitute the derivatives we found for each term:

\[ f'(x) = -3\sec^2(3x)\sin(\tan 3x) + 3\sec^2(3x)\cos(\tan 3x) \]

To simplify the expression, we can factor out the common term \(3\sec^2(3x)\) from both parts:

\[ f'(x) = 3\sec^2(3x)(\cos(\tan 3x) - \sin(\tan 3x)) \]

This is the final differentiated form of the function \(f(x)\).

Matching the Result with Options

Comparing our derived result, \(3\sec^2(3x)(\cos(\tan 3x) - \sin(\tan 3x))\), with the provided options, we find that it matches the option which states \(3\sec^2 3x(\cos(\tan 3x) - \sin(\tan 3x))\).