Calculus Examples

$a x e^{- b x}$

Step 1

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

$a \frac{d}{d x} [x e^{- b x}]$

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

$a (x \frac{d}{d x} [e^{- b x}] + e^{- b x} \frac{d}{d x} [x])$

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 $g (x) = - b x$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

$a (x (e^{- b x} \frac{d}{d x} [- b x]) + e^{- b x} \frac{d}{d x} [x])$

Step 4

Differentiate.

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Step 4.1

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

$a (x (e^{- b x} (- b \frac{d}{d x} [x])) + e^{- b x} \frac{d}{d x} [x])$

Step 4.2

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

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

Step 4.3

Multiply $- 1$ by $1$ .

$a (x (e^{- b x} (- b)) + e^{- b x} \frac{d}{d x} [x])$

Step 4.4

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

$a (x (e^{- b x} (- b)) + e^{- b x} \cdot 1)$

Step 4.5

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

$a (x (e^{- b x} (- b)) + e^{- b x})$

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

$a (x (e^{- b x} (- b))) + a e^{- b x}$

Step 5.2

Reorder terms.

$- e^{- b x} a b x + e^{- b x} a$

Step 5.3

Reorder factors in $- e^{- b x} a b x + e^{- b x} a$ .

$- a b x e^{- b x} + a e^{- b x}$