Question Statement

Find the derivative of $y=\sin 2x\cos 3x$ with respect to $x$.

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Background and Explanation

This problem involves differentiating a product of two composite trigonometric functions. You will need to apply the product rule combined with the chain rule to handle the coefficients inside the sine and cosine functions.

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Solution

We are given the function: $y=\sin 2x\cos 3x$

Since $y$ is a product of two functions, we apply the product rule. Let $u=\sin 2x$ and $v=\cos 3x$. The product rule states: $\frac{d}{dx}(uv)=\frac{du}{dx}\cdot v+u\cdot \frac{dv}{dx}$

Differentiating with respect to $x$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{d}{dx}(\sin 2x)\cdot \cos 3x+\sin 2x\cdot \frac{d}{dx}(\cos 3x)\\ & =\cos 2x\cdot 2\cdot \cos 3x+\sin 2x\cdot (-\sin 3x)\cdot 3\\ & =2\cos 2x\cos 3x-3\sin 2x\sin 3x\end{array}$

Step-by-step reasoning:

• In the first term, we differentiate $\sin 2x$ using the chain rule: $\frac{d}{dx}(\sin 2x)=\cos 2x\cdot 2=2\cos 2x$
• In the second term, we differentiate $\cos 3x$ using the chain rule: $\frac{d}{dx}(\cos 3x)=-\sin 3x\cdot 3=-3\sin 3x$
• Multiplying through and combining terms gives the final result

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Key Formulas or Methods Used

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

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Summary of Steps

1. Identify $y$ as a product of two functions: $\sin 2x$ and $\cos 3x$
2. Apply the product rule: $\frac{dy}{dx}=\frac{d}{dx}(\sin 2x)\cdot \cos 3x+\sin 2x\cdot \frac{d}{dx}(\cos 3x)$
3. Use the chain rule to find $\frac{d}{dx}(\sin 2x)=2\cos 2x$
4. Use the chain rule to find $\frac{d}{dx}(\cos 3x)=-3\sin 3x$
5. Substitute these derivatives back into the product rule expression
6. Simplify to obtain the final answer: $2\cos 2x\cos 3x-3\sin 2x\sin 3x$