$f(x)=\sin 2 x \cos x$

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Find limits involving trigonometric functions

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f(x)=\sin 2 x \cos x

Apply Product Rule: We need to find the derivative of the function $f(x) = \sin(2x)\cos(x)$ . To do this, we will use the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
Define Functions: Let's denote the first function as $u(x) = \sin(2x)$ and the second function as $v(x) = \cos(x)$ . According to the product rule, the derivative of $f(x)$ with respect to $x$ is $f'(x) = u'(x)v(x) + u(x)v'(x)$ .
Find Derivative of $u(x)$ : Now we need to find the derivative of $u(x) = \sin(2x)$ . The derivative of $\sin(2x)$ with respect to $x$ is $2\cos(2x)$ , because the derivative of $\sin(ax)$ with respect to $x$ is $a\cos(ax)$ , where $a$ is a constant.
Find Derivative of $v(x)$ : Next, we need to find the derivative of $v(x) = \cos(x)$ . The derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$ , because the derivative of $\cos(x)$ is $-\sin(x)$ .
Apply Derivatives to Product Rule: Now we can apply the derivatives we found to the product rule. We have $u'(x) = 2\cos(2x)$ and $v'(x) = -\sin(x)$ . Plugging these into the product rule formula, we get $f'(x) = (2\cos(2x))(\cos(x)) + (\sin(2x))(-\sin(x))$ .
Simplify Expression: Simplify the expression for $f'(x)$ by multiplying the terms. This gives us $f'(x) = 2\cos(2x)\cos(x) - \sin(2x)\sin(x)$ .
Final Answer: The expression for $f'(x)$ is now simplified, and we have found the derivative of the function $f(x) = \sin(2x)\cos(x)$ . The final answer is $f'(x) = 2\cos(2x)\cos(x) - \sin(2x)\sin(x)$ .