Find $\frac{d y}{d x}$ for $y=\frac{\cos (2 x)}{5 e^{3 x}}$ \[ \frac{d y}{d x}= \]
Solution
Step 1 ：We are given the function \(y=\frac{\cos (2 x)}{5 e^{3 x}}\) and we are asked to find its derivative with respect to \(x\).
Step 2 ：To solve this problem, we need to use the quotient rule for differentiation. The quotient rule states that the derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\).
Step 3 ：In this case, \(f(x) = \cos(2x)\) and \(g(x) = 5e^{3x}\).
Step 4 ：We first find the derivatives of \(f(x)\) and \(g(x)\). The derivative of \(f(x) = \cos(2x)\) is \(f'(x) = -2\sin(2x)\). The derivative of \(g(x) = 5e^{3x}\) is \(g'(x) = 15e^{3x}\).
Step 5 ：Now that we have \(f'(x)\) and \(g'(x)\), we can substitute these into the quotient rule to find \(\frac{dy}{dx}\).
Step 6 ：Substituting \(f'(x)\), \(g(x)\), \(f(x)\), and \(g'(x)\) into the quotient rule gives us \(\frac{dy}{dx} = \frac{-10e^{3x}\sin(2x) - 15e^{3x}\cos(2x)}{25e^{6x}}\).
Step 7 ：\(\boxed{\frac{d y}{d x} = \frac{-10e^{3x}\sin(2x) - 15e^{3x}\cos(2x)}{25e^{6x}}}\) is the derivative of the given function with respect to \(x\).
