Find the derivative \( y=\frac{e^{2 x+1}}{x} \) a) \( y^{\prime}=\frac{2 x e^{2 x+1}}{x^{2}} \) b) \( y^{\prime}=\frac{-e^{2 x+1}}{x^{2}} \) c) \( \frac{e^{2 x+1}(2 x-1)}{x^{2}} \) d) \( y^{\prime}=\frac{2 e^{2 x+1}-x e^{2 x}}{x^{2}} \)

To find the derivative of the function \( y = \frac{e^{2x+1}}{x} \), we need to apply the quotient rule, which states that if you have a function expressed as \( \frac{u}{v} \), then the derivative \( y' \) is given by: \[ y' = \frac{u'v - uv'}{v^2} \] Here, \( u = e^{2x + 1} \) and \( v = x \). First, we need to calculate the derivatives \( u' \) and \( v' \): 1. \( u' = \frac{d}{dx}(e^{2x+1}) = e^{2x+1} \cdot 2 \) (using the chain rule) 2. \( v' = \frac{d}{dx}(x) = 1 \) Now, applying the quotient rule: \[ y' = \frac{(2e^{2x+1})x - (e^{2x+1})(1)}{x^2} \] Simplifying that: \[ y' = \frac{2xe^{2x+1} - e^{2x + 1}}{x^2} \] Factoring out \( e^{2x+1} \): \[ y' = \frac{e^{2x + 1}(2x - 1)}{x^2} \] This matches option c): **Final Answer:** c) \( \frac{e^{2 x+1}(2 x-1)}{x^{2}} \)