Question

find the derivative. ( f(x) = (2e^x + x)(2e^x - x) ); ( f(x) = ) submit answer next item

Explanation:

Step1: Identify the product rule

The function \( f(x)=(2e^{x}+x)(2e^{x}-x) \) is a product of two functions, \( u(x) = 2e^{x}+x \) and \( v(x)=2e^{x}-x \). The product rule states that if \( f(x)=u(x)v(x) \), then \( f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x) \).

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

For \( u(x) = 2e^{x}+x \), the derivative of \( 2e^{x} \) with respect to \( x \) is \( 2e^{x} \) (since the derivative of \( e^{x} \) is \( e^{x} \)) and the derivative of \( x \) with respect to \( x \) is \( 1 \). So, \( u^{\prime}(x)=2e^{x} + 1 \).

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

For \( v(x)=2e^{x}-x \), the derivative of \( 2e^{x} \) with respect to \( x \) is \( 2e^{x} \) and the derivative of \( -x \) with respect to \( x \) is \( - 1 \). So, \( v^{\prime}(x)=2e^{x}-1 \).

Step4: Apply the product rule

\[

$$\begin{align*} f^{\prime}(x)&=(2e^{x} + 1)(2e^{x}-x)+(2e^{x}+x)(2e^{x}-1)\\ &=(2e^{x})(2e^{x})-2e^{x}\cdot x+2e^{x}-x+(2e^{x})(2e^{x})-2e^{x}+2e^{x}\cdot x - x\\ &=4e^{2x}-2xe^{x}+2e^{x}-x + 4e^{2x}-2e^{x}+2xe^{x}-x\\ &=(4e^{2x}+4e^{2x})+(-2xe^{x}+2xe^{x})+(2e^{x}-2e^{x})+(-x - x)\\ &=8e^{2x}-2x \end{align*}$$

\]
(Alternatively, we can first expand \( f(x)=(2e^{x})^{2}-x^{2}=4e^{2x}-x^{2} \), then take the derivative. The derivative of \( 4e^{2x} \) is \( 4\times2e^{2x}=8e^{2x} \) (using the chain rule: derivative of \( e^{u} \) is \( e^{u}u^{\prime} \), here \( u = 2x \), \( u^{\prime}=2 \)) and the derivative of \( -x^{2} \) is \( - 2x \). So \( f^{\prime}(x)=8e^{2x}-2x \))

Answer:

\( 8e^{2x}-2x \)