Question

1. if ( f(x) = \frac{e^{2x}}{2x} ), then ( f(x) = )

(a) ( 1 )
(b) ( \frac{e^{2x}(1 - 2x)}{2x^2} )
(c) ( e^{2x} )
(d) ( \frac{e^{2x}(2x + 1)}{x^2} )
(e) ( \frac{e^{2x}(2x - 1)}{2x^2} )

Explanation:

Step1: Identify the rule

We use the quotient rule for differentiation. The quotient rule states that if \( f(x)=\frac{u(x)}{v(x)} \), then \( f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{[v(x)]^{2}} \). Here, \( u(x) = e^{2x} \) and \( v(x)=2x \).

Step2: Find \( u^{\prime}(x) \)

To find the derivative of \( u(x)=e^{2x} \), we use the chain rule. The chain rule states that if \( y = e^{g(x)} \), then \( y^{\prime}=e^{g(x)}\cdot g^{\prime}(x) \). For \( g(x) = 2x \), \( g^{\prime}(x)=2 \). So, \( u^{\prime}(x)=2e^{2x} \).

Step3: Find \( v^{\prime}(x) \)

The derivative of \( v(x) = 2x \) with respect to \( x \) is \( v^{\prime}(x)=2 \).

Step4: Apply the quotient rule

Substitute \( u(x) \), \( u^{\prime}(x) \), \( v(x) \), and \( v^{\prime}(x) \) into the quotient rule formula:
\[

$$\begin{align*} f^{\prime}(x)&=\frac{(2e^{2x})(2x)-e^{2x}(2)}{(2x)^{2}}\\ &=\frac{4xe^{2x}-2e^{2x}}{4x^{2}}\\ &=\frac{2e^{2x}(2x - 1)}{4x^{2}}\\ &=\frac{e^{2x}(2x - 1)}{2x^{2}} \end{align*}$$

\]

Answer:

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