Question Statement

Find the derivative $\frac{dy}{dx}$ of the function:

$y=e^{-3x}\cos x$

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Background and Explanation

This problem involves differentiating a product of two functions: the exponential function $e^{-3x}$ and the trigonometric function $\cos x$. You will need to apply the product rule combined with the chain rule to handle the composite exponential term.

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Solution

Finding $\frac{dy}{dx}$

We are given: $y=e^{-3x}\cos x$

This function is a product of two distinct functions:

• $u=e^{-3x}$ (exponential function with chain rule required)
• $v=\cos x$ (trigonometric function)

To differentiate, we apply the product rule: $\frac{d}{dx}(uv)=\frac{du}{dx}\cdot v+u\cdot \frac{dv}{dx}$

Setting up the differentiation:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{d}{dx}\left(e^{-3x}\right)\cos x+e^{-3x}\cdot \frac{d}{dx}(\cos x)\\ & =e^{-3x}\cdot (-3)\cos x+e^{-3x}(-\sin x)\\ & =e^{-3x}(-3\cos x-\sin x)\\ \frac{dy}{dx}& =-e^{-3x}(3\cos x+\sin x)\end{array}$

Step-by-step reasoning:

1. First line: Apply the product rule structure—derivative of the first function ($e^{-3x}$) times the second function ($\cos x$), plus the first function times the derivative of the second function ($\cos x$).

2. Second line: Evaluate each derivative:

   • $\frac{d}{dx}(e^{-3x})=e^{-3x}\cdot (-3)=-3e^{-3x}$ (using the chain rule: derivative of $-3x$ is $-3$)
   • $\frac{d}{dx}(\cos x)=-\sin x$

3. Third line: Factor out the common term $e^{-3x}$ from both terms to simplify the expression.

4. Fourth line: Factor out $-1$ from the parentheses to present the final answer in its most compact form: $-e^{-3x}(3\cos x+\sin x)$.

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Derivative of exponential: $\frac{d}{dx}(e^{ax})=a{e}^{ax}$
• Derivative of cosine: $\frac{d}{dx}(\cos x)=-\sin x$

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Summary of Steps

1. Identify $y=e^{-3x}\cos x$ as a product of two functions requiring the product rule.
2. Differentiate $e^{-3x}$ using the chain rule to obtain $-3e^{-3x}$.
3. Differentiate $\cos x$ to obtain $-\sin x$.
4. Apply the product rule formula: $\frac{dy}{dx}=(-3e^{-3x})\cos x+e^{-3x}(-\sin x)$.
5. Simplify by factoring out $e^{-3x}$: $\frac{dy}{dx}=e^{-3x}(-3\cos x-\sin x)$.
6. Factor out $-1$ to obtain the final result: $\frac{dy}{dx}=-e^{-3x}(3\cos x+\sin x)$.