Calculus Examples

$4 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}$

Step 1

By the Sum Rule, the derivative of $4 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}$ with respect to $x$ is $\frac{d}{d x} [4 x^{2} e^{- x^{2}}] + \frac{d}{d x} [- 2 e^{- x^{2}}]$ .

$\frac{d}{d x} [4 x^{2} e^{- x^{2}}] + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2

Evaluate $\frac{d}{d x} [4 x^{2} e^{- x^{2}}]$ .

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Step 2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2} e^{- x^{2}}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2} e^{- x^{2}}]$ .

$4 \frac{d}{d x} [x^{2} e^{- x^{2}}] + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.2

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

$4 (x^{2} \frac{d}{d x} [e^{- x^{2}}] + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.3

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 $g (x) = - x^{2}$ .

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Step 2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x^{2}$ .

$4 (x^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- x^{2}]) + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$4 (x^{2} (e^{u_{1}} \frac{d}{d x} [- x^{2}]) + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.3.3

Replace all occurrences of $u_{1}$ with $- x^{2}$ .

$4 (x^{2} (e^{- x^{2}} \frac{d}{d x} [- x^{2}]) + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$4 (x^{2} (e^{- x^{2}} (- \frac{d}{d x} [x^{2}])) + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.5

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

$4 (x^{2} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.6

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

$4 (x^{2} (e^{- x^{2}} (- (2 x))) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.7

Multiply $2$ by $- 1$ .

$4 (x^{2} (e^{- x^{2}} (- 2 x)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.8

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.8.1

Move $x$ .

$4 (x \cdot x^{2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.8.2

Multiply $x$ by $x^{2}$ .

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Step 2.8.2.1

Raise $x$ to the power of $1$ .

$4 (x^{1} x^{2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 (x^{1 + 2} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.8.3

Add $1$ and $2$ .

$4 (x^{3} (e^{- x^{2}} \cdot (- 2)) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 2.9

Move $- 2$ to the left of $e^{- x^{2}}$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + \frac{d}{d x} [- 2 e^{- x^{2}}]$

Step 3

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

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Step 3.1

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

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 \frac{d}{d x} [e^{- x^{2}}]$

Step 3.2

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 $g (x) = - x^{2}$ .

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Step 3.2.1

To apply the Chain Rule, set $u_{2}$ as $- x^{2}$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x^{2}])$

Step 3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (e^{u_{2}} \frac{d}{d x} [- x^{2}])$

Step 3.2.3

Replace all occurrences of $u_{2}$ with $- x^{2}$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (e^{- x^{2}} \frac{d}{d x} [- x^{2}])$

Step 3.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (e^{- x^{2}} (- \frac{d}{d x} [x^{2}]))$

Step 3.4

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

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (e^{- x^{2}} (- (2 x)))$

Step 3.5

Multiply $2$ by $- 1$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) - 2 (e^{- x^{2}} (- 2 x))$

Step 3.6

Multiply $- 2$ by $- 2$ .

$4 (x^{3} (- 2 e^{- x^{2}}) + e^{- x^{2}} (2 x)) + 4 e^{- x^{2}} x$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$4 (x^{3} (- 2 e^{- x^{2}})) + 4 (e^{- x^{2}} (2 x)) + 4 e^{- x^{2}} x$

Step 4.2

Combine terms.

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Step 4.2.1

Multiply $- 2$ by $4$ .

$- 8 (x^{3} (e^{- x^{2}})) + 4 (e^{- x^{2}} (2 x)) + 4 e^{- x^{2}} x$

Step 4.2.2

Multiply $2$ by $4$ .

$- 8 x^{3} e^{- x^{2}} + 8 (e^{- x^{2}} (x)) + 4 e^{- x^{2}} x$

Step 4.2.3

Add $8 e^{- x^{2}} x$ and $4 e^{- x^{2}} x$ .

$- 8 x^{3} e^{- x^{2}} + 12 e^{- x^{2}} x$

Step 4.3

Reorder terms.

$- 8 e^{- x^{2}} x^{3} + 12 e^{- x^{2}} x$

Step 4.4

Reorder factors in $- 8 e^{- x^{2}} x^{3} + 12 e^{- x^{2}} x$ .

$- 8 x^{3} e^{- x^{2}} + 12 x e^{- x^{2}}$