Calculus Examples

$e^{- 5 x^{2}}$

Find the first derivative.

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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 $g (x) = - 5 x^{2}$ .

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To apply the Chain Rule, set $u$ as $- 5 x^{2}$ .

$f' (x) = \frac{d}{d u} (e^{u}) \frac{d}{d x} (- 5 x^{2})$

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

$f' (x) = e^{u} \frac{d}{d x} (- 5 x^{2})$

Replace all occurrences of $u$ with $- 5 x^{2}$ .

$f' (x) = e^{- 5 x^{2}} \frac{d}{d x} (- 5 x^{2})$

Differentiate.

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Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$f' (x) = e^{- 5 x^{2}} (- 5 \frac{d}{d x} x^{2})$

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

$f' (x) = e^{- 5 x^{2}} (- 5 (2 x))$

Multiply $2$ by $- 5$ .

$f' (x) = e^{- 5 x^{2}} (- 10 x)$

Simplify.

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Reorder the factors of $e^{- 5 x^{2}} (- 10 x)$ .

$f' (x) = - 10 e^{- 5 x^{2}} x$

Reorder factors in $- 10 e^{- 5 x^{2}} x$ .

$f' (x) = - 10 x e^{- 5 x^{2}}$

Find the second derivative.

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Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x e^{- 5 x^{2}}$ with respect to $x$ is $- 10 \frac{d}{d x} [x e^{- 5 x^{2}}]$ .

$f'' (x) = - 10 \frac{d}{d x} (x e^{- 5 x^{2}})$

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 $g (x) = e^{- 5 x^{2}}$ .

$f'' (x) = - 10 (x \frac{d}{d x} (e^{- 5 x^{2}}) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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 $g (x) = - 5 x^{2}$ .

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To apply the Chain Rule, set $u$ as $- 5 x^{2}$ .

$f'' (x) = - 10 (x (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- 5 x^{2})) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x (e^{u} \frac{d}{d x} (- 5 x^{2})) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

Replace all occurrences of $u$ with $- 5 x^{2}$ .

$f'' (x) = - 10 (x (e^{- 5 x^{2}} \frac{d}{d x} (- 5 x^{2})) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

Differentiate.

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Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{2}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 10 (x (e^{- 5 x^{2}} (- 5 \frac{d}{d x} x^{2})) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x (e^{- 5 x^{2}} (- 5 (2 x))) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

Multiply $2$ by $- 5$ .

$f'' (x) = - 10 (x (e^{- 5 x^{2}} (- 10 x)) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x \cdot x (e^{- 5 x^{2}} \cdot (- 10)) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x \cdot x (e^{- 5 x^{2}} \cdot (- 10)) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x^{1 + 1} (e^{- 5 x^{2}} \cdot (- 10)) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

Simplify the expression.

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Add $1$ and $1$ .

$f'' (x) = - 10 (x^{2} (e^{- 5 x^{2}} \cdot (- 10)) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

Move $- 10$ to the left of $e^{- 5 x^{2}}$ .

$f'' (x) = - 10 (x^{2} (- 10 \cdot e^{- 5 x^{2}}) + e^{- 5 x^{2}} \frac{d}{d x} (x))$

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

$f'' (x) = - 10 (x^{2} (- 10 e^{- 5 x^{2}}) + e^{- 5 x^{2}} \cdot 1)$

Multiply $e^{- 5 x^{2}}$ by $1$ .

$f'' (x) = - 10 (x^{2} (- 10 e^{- 5 x^{2}}) + e^{- 5 x^{2}})$

Simplify.

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Apply the distributive property.

$f'' (x) = - 10 (x^{2} (- 10 e^{- 5 x^{2}})) - 10 e^{- 5 x^{2}}$

Multiply $- 10$ by $- 10$ .

$f'' (x) = 100 x^{2} e^{- 5 x^{2}} - 10 e^{- 5 x^{2}}$

Reorder terms.

$f'' (x) = 100 e^{- 5 x^{2}} x^{2} - 10 e^{- 5 x^{2}}$

Reorder factors in $100 e^{- 5 x^{2}} x^{2} - 10 e^{- 5 x^{2}}$ .

$f'' (x) = 100 x^{2} e^{- 5 x^{2}} - 10 e^{- 5 x^{2}}$