Find the 3rd Derivative xe^x

$x e^{x}$

Step 1

Find the first derivative.

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x)$

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

$f' (x) = x e^{x} + e^{x} \frac{d}{d x} (x)$

Differentiate using the Power Rule.

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

$f' (x) = x e^{x} + e^{x} \cdot 1$

Multiply $e^{x}$ by $1$ .

$f' (x) = x e^{x} + e^{x}$

Step 2

Find the second derivative.

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

$f'' (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (e^{x})$

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

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f'' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (e^{x})$

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

$f'' (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (e^{x})$

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

$f'' (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (e^{x})$

Multiply $e^{x}$ by $1$ .

$f'' (x) = x e^{x} + e^{x} + \frac{d}{d x} (e^{x})$

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

$f'' (x) = x e^{x} + e^{x} + e^{x}$

Simplify.

Add $e^{x}$ and $e^{x}$ .

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

Reorder terms.

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

Reorder factors in $e^{x} x + 2 e^{x}$ .

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

Step 3

Find the third derivative.

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

$f''' (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (2 e^{x})$

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

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f''' (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (2 e^{x})$

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

$f''' (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (2 e^{x})$

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

$f''' (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (2 e^{x})$

Multiply $e^{x}$ by $1$ .

$f''' (x) = x e^{x} + e^{x} + \frac{d}{d x} (2 e^{x})$

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

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

$f''' (x) = x e^{x} + e^{x} + 2 \frac{d}{d x} (e^{x})$

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

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

Simplify.

Add $e^{x}$ and $2 e^{x}$ .

$f''' (x) = x e^{x} + 3 e^{x}$

Reorder terms.

$f''' (x) = e^{x} x + 3 e^{x}$

Reorder factors in $e^{x} x + 3 e^{x}$ .

$f''' (x) = x e^{x} + 3 e^{x}$

Step 4

Find the fourth derivative.

By the Sum Rule, the derivative of $x e^{x} + 3 e^{x}$ with respect to $x$ is $\frac{d}{d x} [x e^{x}] + \frac{d}{d x} [3 e^{x}]$ .

$f^{4} (x) = \frac{d}{d x} (x e^{x}) + \frac{d}{d x} (3 e^{x})$

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

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{x}$ .

$f^{4} (x) = x \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (3 e^{x})$

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

$f^{4} (x) = x e^{x} + e^{x} \frac{d}{d x} (x) + \frac{d}{d x} (3 e^{x})$

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

$f^{4} (x) = x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} (3 e^{x})$

Multiply $e^{x}$ by $1$ .

$f^{4} (x) = x e^{x} + e^{x} + \frac{d}{d x} (3 e^{x})$

Evaluate $\frac{d}{d x} [3 e^{x}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{x}]$ .

$f^{4} (x) = x e^{x} + e^{x} + 3 \frac{d}{d x} (e^{x})$

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

$f^{4} (x) = x e^{x} + e^{x} + 3 e^{x}$

Simplify.

Add $e^{x}$ and $3 e^{x}$ .

$f^{4} (x) = x e^{x} + 4 e^{x}$

Reorder terms.

$f^{4} (x) = e^{x} x + 4 e^{x}$

Reorder factors in $e^{x} x + 4 e^{x}$ .

$f^{4} (x) = x e^{x} + 4 e^{x}$

Find the 3rd Derivative xe^x

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