Question

1. find the derivative.

f(x)=(2e^x + x)(2e^x - x);
f(x)=

Explanation:

Step1: Use the difference - of - squares formula

First, expand $(2e^{x}+x)(2e^{x}-x)$ using the formula $(a + b)(a - b)=a^{2}-b^{2}$. Here $a = 2e^{x}$ and $b=x$, so $f(x)=(2e^{x})^{2}-x^{2}=4e^{2x}-x^{2}$.

Step2: Apply the sum - rule of derivatives

The sum - rule states that if $y = u - v$, then $y^\prime=u^\prime - v^\prime$. Let $u = 4e^{2x}$ and $v=x^{2}$.

Step3: Differentiate $u = 4e^{2x}$

Using the chain - rule. If $y = e^{u}$ and $u = 2x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. For $y = 4e^{u}$ with $u = 2x$, $\frac{du}{dx}=2$ and $\frac{dy}{du}=4e^{u}$, so $\frac{d}{dx}(4e^{2x})=4e^{2x}\cdot2 = 8e^{2x}$.

Step4: Differentiate $v=x^{2}$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $n = 2$, $\frac{d}{dx}(x^{2})=2x$.

Step5: Find $f^\prime(x)$

By the sum - rule $f^\prime(x)=\frac{d}{dx}(4e^{2x})-\frac{d}{dx}(x^{2})$. So $f^\prime(x)=8e^{2x}-2x$.

Answer:

$8e^{2x}-2x$