find $\\frac{d}{dx}\\left\\frac{e^{2x}}{2x}\\right$.\nchoose 1 answer:\na $\\frac{e^{2x}(2x…

find $\\frac{d}{dx}\\left\\frac{e^{2x}}{2x}\\right$.\nchoose 1 answer:\na $\\frac{e^{2x}(2x - 1)}{2x^{2}}$\nb $\\frac{e^{2x}(x - 1)}{x}$\nc $\\frac{e^{2x}}{2}$\nd $e^{2x}$

find $\\frac{d}{dx}\\left\\frac{e^{2x}}{2x}\\right$.\nchoose 1 answer:\na $\\frac{e^{2x}(2x - 1)}{2x^{2}}$\nb $\\frac{e^{2x}(x - 1)}{x}$\nc $\\frac{e^{2x}}{2}$\nd $e^{2x}$

Answer

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (y = \frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here, (u = e^{2x}), (u^\prime=2e^{2x}) (by the chain rule: if (y = e^{f(x)}), (y^\prime=f^\prime(x)e^{f(x)}), and (f(x) = 2x), (f^\prime(x)=2)), and (v = 2x), (v^\prime = 2).

Step2: Substitute into the quotient rule formula

[ \begin{align*} \frac{d}{dx}\left(\frac{e^{2x}}{2x}\right)&=\frac{(2e^{2x})\times(2x)-e^{2x}\times2}{(2x)^{2}}\ &=\frac{4xe^{2x}- 2e^{2x}}{4x^{2}}\ &=\frac{2e^{2x}(2x - 1)}{4x^{2}}\ &=\frac{e^{2x}(2x - 1)}{2x^{2}} \end{align*} ]

Answer:

A. (\frac{e^{2x}(2x - 1)}{2x^{2}})