Calculus Examples

$(x^{2} - 3) e^{- x}$

Step 1

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 $f (x) = x^{2} - 3$ and $g (x) = e^{- x}$ .

$(x^{2} - 3) \frac{d}{d x} [e^{- x}] + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 2

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 $g (x) = - x$ .

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Step 2.1

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

$(x^{2} - 3) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 2.2

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

$(x^{2} - 3) (e^{u} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 2.3

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

$(x^{2} - 3) (e^{- x} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3

Differentiate.

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Step 3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$(x^{2} - 3) e^{- x} (- \frac{d}{d x} [x]) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3.2

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

$(x^{2} - 3) e^{- x} (- 1 \cdot 1) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $- 1$ by $1$ .

$(x^{2} - 3) e^{- x} \cdot - 1 + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3.3.2

Move $- 1$ to the left of $(x^{2} - 3) e^{- x}$ .

$- 1 \cdot ((x^{2} - 3) e^{- x}) + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3.3.3

Rewrite $- 1 (x^{2} - 3)$ as $- (x^{2} - 3)$ .

$- (x^{2} - 3) e^{- x} + e^{- x} \frac{d}{d x} [x^{2} - 3]$

Step 3.4

By the Sum Rule, the derivative of $x^{2} - 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3]$ .

$- (x^{2} - 3) e^{- x} + e^{- x} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3])$

Step 3.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- (x^{2} - 3) e^{- x} + e^{- x} (2 x + \frac{d}{d x} [- 3])$

Step 3.6

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$- (x^{2} - 3) e^{- x} + e^{- x} (2 x + 0)$

Step 3.7

Add $2 x$ and $0$ .

$- (x^{2} - 3) e^{- x} + e^{- x} (2 x)$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$(- x^{2} - - 3) e^{- x} + e^{- x} (2 x)$

Step 4.2

Apply the distributive property.

$- x^{2} e^{- x} - - 3 e^{- x} + e^{- x} (2 x)$

Step 4.3

Multiply $- 1$ by $- 3$ .

$- x^{2} e^{- x} + 3 e^{- x} + e^{- x} (2 x)$

Step 4.4

Reorder terms.

$- e^{- x} x^{2} + 2 e^{- x} x + 3 e^{- x}$

Step 4.5

Reorder factors in $- e^{- x} x^{2} + 2 e^{- x} x + 3 e^{- x}$ .

$- x^{2} e^{- x} + 2 x e^{- x} + 3 e^{- x}$