Use chain rule: \(rac{dy}{dx} = \cos(3x) \cdot rac{d}{dx}(3x) = \cos(3x) \cdot 3 = 3\cos(3x)\).

["# Understanding the Chain Rule: Derivative of ( \frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) )", "When learning calculus, one of the most powerful tools for finding derivatives is the chain rule. This fundamental principle allows us to differentiate composite functions — functions nested inside other functions — with precision and confidence.", "In this article, we’ll explore how the chain rule applies to the derivative of ( y = \cos(3x) ), following the structure:\n[\n\frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) = \cos(3x) \cdot 3 = 3\cos(3x)\n]", "## What is the Chain Rule?", "The chain rule states that if ( y = f(u) ) and ( u = g(x) ), then the derivative of ( y ) with respect to ( x ) is:\n[\n\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}\n]\nIn other words, differentiate the outer function first, treating the inner function as a single variable, then multiply by the derivative of the inner function.", "## Applying the Chain Rule: Step-by-Step Example", "Let’s analyze the function:\n[\ny = \cos(3x)\n]", "Here, the inner function is ( u = 3x ), and the outer function is ( y = \cos(u) ).", "### Step 1: Differentiate the outer function\nWe know that the derivative of ( \cos(u) ) with respect to ( u ) is:\n[\n\frac{dy}{du} = -\sin(u)\n]\nSubstituting back ( u = 3x ), we get:\n[\n\frac{d}{dx}[\cos(3x)] = -\sin(3x) \cdot \frac{d}{dx}(3x)\n]", "### Step 2: Differentiate the inner function\nThe derivative of ( u = 3x ) with respect to ( x ) is simply:\n[\n\frac{du}{dx} = 3\n]", "### Step 3: Combine using the chain rule\nNow multiply the two derivatives:\n[\n\frac{dy}{dx} = \left(-\sin(3x)\right) \cdot 3 = -3\sin(3x)\n]", "Wait — if we re-evaluate the chain rule correctly, remember the derivative of ( \cos(u) ) is ( -\sin(u) ), but sometimes sign varies depending on context — here, our direct computation confirms:\n[\n\frac{d}{dx}[\cos(3x)] = -\sin(3x) \cdot 3 = -3\sin(3x)\n]", "However, if the original expression was intended as a product or alternate form such as ( \frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) ), this interpretation is incomplete or misleading without proper function structure. The correct and clean evaluation is:", "[\n\frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) = \cos(3x) \cdot 3 = 3\cos(3x)\n]", "But note: this only directly applies if ( y = \cos(u) ) and ( u = 3x ), so yes, the chain rule gives:\n[\n\frac{dy}{dx} = -\sin(3x) \cdot 3 = -3\sin(3x)\n]", "So, interpretation depends on how ( y ) is defined. Suppose instead the function is written in product or more complex form — that’s where chain rule shines. But in its simplest form, let’s clarify:", "If ( y = \cos(3x) ), derivative via chain rule is:\n[\n\frac{dy}{dx} = -\sin(3x) \cdot 3 = -3\sin(3x)\n]", "But if the expression:\n[\n\frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x)\n]\nis meant to reflect ( y ) as a product where one factor is ( \cos(3x) ) and its derivative is multiplied externally, this phrasing can mislead. The proper interpretation is:", "### Correct Derivation Clarified:\nLet ( y = \cos(3x) ). Then,\n[\n\frac{dy}{dx} = -\sin(3x) \cdot \frac{d}{dx}(3x) = -\sin(3x) \cdot 3 = -3\sin(3x)\n]", "However, if someone mistakenly applies the chain rule assuming a different structure — such as differentiating a composite where outer function is cosine and inner is 3x — then yes:\n[\n\frac{dy}{dx} = \cos(u), \quad u = 3x \Rightarrow \frac{dy}{dx} = -\sin(u) \cdot 3 = -3\sin(3x)\n]", "If the expression ( \frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) ) is taken literally with no product or composition originally presented, it’s inconsistent. But when interpreted properly, the chain rule properly applied to ( y = \cos(3x) ) yields:\n[\n\boxed{ \frac{dy}{dx} = -3\sin(3x) }\n]", "### Why This Matters: Chain Rule in Action", "Using the chain rule correctly transforms complex functional derivatives into manageable components. Whether ( y = \sin(5x) ), ( y = e^{2x} \sin(x) ), or ( y = \cos(4x) ), the chain rule remains central: identify the outer function and inner function, differentiate accordingly, and multiply.", "Memorizing the pattern:\n[\n\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx}, \quad u = 3x \Rightarrow \frac{du}{dx} = 3\n]\nSo:\n[\n\frac{dy}{dx} = -\sin(3x) \cdot 3 = -3\sin(3x)\n]", "### Final Thoughts", "The chain rule is indispensable in calculus. Whether differentiating simple trigonometric composites or multi-layered functions, mastering this tool builds confidence and precision in derivatives.", "Keywords: chain rule, derivative of cosine, product rule vs chain rule, differentiation chain rule, ( \frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) ), mathematical chain rule explained, calculus step-by-step guide.", "Meta Description:\nLearn how the chain rule works using ( \frac{dy}{dx} = \cos(3x) \cdot \frac{d}{dx}(3x) ). Step-by-step breakdown shows derivative equals ( -3\sin(3x) ) with clear chain rule application.", "---", "Understanding the chain rule opens doors to more advanced calculus and real-world applications in physics, engineering, and optimization. Keep practicing composite functions — it pays off!"]