Find the Derivative - d/d@VAR f(x)=6+2e^(-2x)

$f (x) = 6 + 2 e^{- 2 x}$

Step 1

Differentiate.

By the Sum Rule, the derivative of $6 + 2 e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [6] + \frac{d}{d x} [2 e^{- 2 x}]$ .

$\frac{d}{d x} [6] + \frac{d}{d x} [2 e^{- 2 x}]$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [2 e^{- 2 x}]$

Step 2

Evaluate $\frac{d}{d x} [2 e^{- 2 x}]$ .

Since $2$ is constant with respect to $x$ , the derivative of $2 e^{- 2 x}$ with respect to $x$ is $2 \frac{d}{d x} [e^{- 2 x}]$ .

$0 + 2 \frac{d}{d x} [e^{- 2 x}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 2 x$ .

To apply the Chain Rule, set $u$ as $- 2 x$ .

$0 + 2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 2 x])$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$0 + 2 (e^{u} \frac{d}{d x} [- 2 x])$

Replace all occurrences of $u$ with $- 2 x$ .

$0 + 2 (e^{- 2 x} \frac{d}{d x} [- 2 x])$

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 + 2 (e^{- 2 x} (- 2 \frac{d}{d x} [x]))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$0 + 2 (e^{- 2 x} (- 2 \cdot 1))$

Multiply $- 2$ by $1$ .

$0 + 2 (e^{- 2 x} \cdot - 2)$

Move $- 2$ to the left of $e^{- 2 x}$ .

$0 + 2 (- 2 \cdot e^{- 2 x})$

Multiply $- 2$ by $2$ .

$0 - 4 e^{- 2 x}$

Step 3

Subtract $4 e^{- 2 x}$ from $0$ .

$- 4 e^{- 2 x}$