Calculus Examples

$y = e^{- x} + 2 x e^{- x} + x^{2} e^{- x}$

Step 1

Find the first derivative.

Tap for more steps...

Step 1.1

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

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

Step 1.2

Since $e^{- x}$ is constant with respect to $e$ , the derivative of $e^{- x}$ with respect to $e$ is $0$ .

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

Step 1.3

Since $2 x e^{- x}$ is constant with respect to $e$ , the derivative of $2 x e^{- x}$ with respect to $e$ is $0$ .

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

Step 1.4

Since $x^{2} e^{- x}$ is constant with respect to $e$ , the derivative of $x^{2} e^{- x}$ with respect to $e$ is $0$ .

$0 + 0 + 0$

Step 1.5

Combine terms.

Tap for more steps...

Step 1.5.1

Add $0$ and $0$ .

$0 + 0$

Step 1.5.2

Add $0$ and $0$ .

$f' (e) = 0$

Step 2

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

$f'' (e) = 0$