Calculus Examples

$9 x e^{- k x}$

Step 1

Since $9$ is constant with respect to $x$ , the derivative of $9 x e^{- k x}$ with respect to $x$ is $9 \frac{d}{d x} [x e^{- k x}]$ .

$9 \frac{d}{d x} [x e^{- k x}]$

Step 2

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 $g (x) = e^{- k x}$ .

$9 (x \frac{d}{d x} [e^{- k x}] + e^{- k x} \frac{d}{d x} [x])$

Step 3

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 $g (x) = - k x$ .

Tap for more steps...

Step 3.1

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

$9 (x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- k x]) + e^{- k x} \frac{d}{d x} [x])$

Step 3.2

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

$9 (x (e^{u} \frac{d}{d x} [- k x]) + e^{- k x} \frac{d}{d x} [x])$

Step 3.3

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

$9 (x (e^{- k x} \frac{d}{d x} [- k x]) + e^{- k x} \frac{d}{d x} [x])$

Step 4

Differentiate.

Tap for more steps...

Step 4.1

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

$9 (x (e^{- k x} (- k \frac{d}{d x} [x])) + e^{- k x} \frac{d}{d x} [x])$

Step 4.2

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

$9 (x (e^{- k x} (- k \cdot 1)) + e^{- k x} \frac{d}{d x} [x])$

Step 4.3

Multiply $- 1$ by $1$ .

$9 (x (e^{- k x} (- k)) + e^{- k x} \frac{d}{d x} [x])$

Step 4.4

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

$9 (x (e^{- k x} (- k)) + e^{- k x} \cdot 1)$

Step 4.5

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

$9 (x (e^{- k x} (- k)) + e^{- k x})$

Step 5

Simplify.

Tap for more steps...

Step 5.1

Apply the distributive property.

$9 (x (e^{- k x} (- k))) + 9 e^{- k x}$

Step 5.2

Multiply $- 1$ by $9$ .

$- 9 x (e^{- k x} (k)) + 9 e^{- k x}$

Step 5.3

Reorder terms.

$- 9 e^{- k x} k x + 9 e^{- k x}$

Step 5.4

Reorder factors in $- 9 e^{- k x} k x + 9 e^{- k x}$ .

$- 9 k x e^{- k x} + 9 e^{- k x}$