Calculus Examples

$e^{- 1.5 x^{2}}$

Step 1

Write $e^{- 1.5 x^{2}}$ as a function.

$f (x) = e^{- 1.5 x^{2}}$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 $g (x) = - 1.5 x^{2}$ .

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Step 2.1.1.1

To apply the Chain Rule, set $u$ as $- 1.5 x^{2}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 1.5 x^{2}]$

Step 2.1.1.2

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

$e^{u} \frac{d}{d x} [- 1.5 x^{2}]$

Step 2.1.1.3

Replace all occurrences of $u$ with $- 1.5 x^{2}$ .

$e^{- 1.5 x^{2}} \frac{d}{d x} [- 1.5 x^{2}]$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

Since $- 1.5$ is constant with respect to $x$ , the derivative of $- 1.5 x^{2}$ with respect to $x$ is $- 1.5 \frac{d}{d x} [x^{2}]$ .

$e^{- 1.5 x^{2}} (- 1.5 \frac{d}{d x} [x^{2}])$

Step 2.1.2.2

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

$e^{- 1.5 x^{2}} (- 1.5 (2 x))$

Step 2.1.2.3

Multiply $2$ by $- 1.5$ .

$e^{- 1.5 x^{2}} (- 3 x)$

Step 2.1.3

Simplify.

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Step 2.1.3.1

Reorder the factors of $e^{- 1.5 x^{2}} (- 3 x)$ .

$- 3 e^{- 1.5 x^{2}} x$

Step 2.1.3.2

Reorder factors in $- 3 e^{- 1.5 x^{2}} x$ .

$f' (x) = - 3 x e^{- 1.5 x^{2}}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $- 3 x e^{- 1.5 x^{2}}$ .

$- 3 x e^{- 1.5 x^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $- 3 x e^{- 1.5 x^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$- 3 x e^{- 1.5 x^{2}} = 0$

Step 3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$e^{- 1.5 x^{2}} = 0$

Step 3.3

Set $x$ equal to $0$ .

$x = 0$

Step 3.4

Set $e^{- 1.5 x^{2}}$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $e^{- 1.5 x^{2}}$ equal to $0$ .

$e^{- 1.5 x^{2}} = 0$

Step 3.4.2

Solve $e^{- 1.5 x^{2}} = 0$ for $x$ .

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Step 3.4.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- 1.5 x^{2}}) = \ln (0)$

Step 3.4.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 3.4.2.3

There is no solution for $e^{- 1.5 x^{2}} = 0$

No solution

Step 3.5

The final solution is all the values that make $- 3 x e^{- 1.5 x^{2}} = 0$ true.

$x = 0$

Step 4

The values which make the derivative equal to $0$ are $0$ .

$0$

Step 5

After finding the point that makes the derivative $f' (x) = - 3 x e^{- 1.5 x^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = e^{- 1.5 x^{2}}$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = - 3 \cdot - 1 \cdot e^{- 1.5 {(- 1)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $- 3$ by $- 1$ .

$f' (- 1) = 3 \cdot e^{- 1.5 {(- 1)}^{2}}$

Step 6.2.2

Raise $- 1$ to the power of $2$ .

$f' (- 1) = 3 \cdot e^{- 1.5 \cdot 1}$

Step 6.2.3

Multiply $- 1.5$ by $1$ .

$f' (- 1) = 3 \cdot e^{- 1.5}$

Step 6.2.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (- 1) = 3 \cdot \frac{1}{e^{1.5}}$

Step 6.2.5

Combine $3$ and $\frac{1}{e^{1.5}}$ .

$f' (- 1) = \frac{3}{e^{1.5}}$

Step 6.2.6

Replace $e$ with an approximation.

$f' (- 1) = \frac{3}{{2.71828182}^{1.5}}$

Step 6.2.7

Raise $2.71828182$ to the power of $1.5$ .

$f' (- 1) = \frac{3}{4.48168907}$

Step 6.2.8

Divide $3$ by $4.48168907$ .

$f' (- 1) = 0.66939048$

Step 6.2.9

The final answer is $0.66939048$ .

$0.66939048$

Step 6.3

At $x = - 1$ the derivative is $0.66939048$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = - 3 \cdot 1 \cdot e^{- 1.5 {(1)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $- 3$ by $1$ .

$f' (1) = - 3 \cdot e^{- 1.5 {(1)}^{2}}$

Step 7.2.2

One to any power is one.

$f' (1) = - 3 \cdot e^{- 1.5 \cdot 1}$

Step 7.2.3

Multiply $- 1.5$ by $1$ .

$f' (1) = - 3 \cdot e^{- 1.5}$

Step 7.2.4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (1) = - 3 \cdot \frac{1}{e^{1.5}}$

Step 7.2.5

Combine $- 3$ and $\frac{1}{e^{1.5}}$ .

$f' (1) = \frac{- 3}{e^{1.5}}$

Step 7.2.6

Move the negative in front of the fraction.

$f' (1) = - \frac{3}{e^{1.5}}$

Step 7.2.7

Replace $e$ with an approximation.

$f' (1) = - \frac{3}{{2.71828182}^{1.5}}$

Step 7.2.8

Raise $2.71828182$ to the power of $1.5$ .

$f' (1) = - \frac{3}{4.48168907}$

Step 7.2.9

Divide $3$ by $4.48168907$ .

$f' (1) = - 1 \cdot 0.66939048$

Step 7.2.10

Multiply $- 1$ by $0.66939048$ .

$f' (1) = - 0.66939048$

Step 7.2.11

The final answer is $- 0.66939048$ .

$- 0.66939048$

Step 7.3

At $x = 1$ the derivative is $- 0.66939048$ . Since this is negative, the function is decreasing on $(0, \infty)$ .

Decreasing on $(0, \infty)$ since $f' (x) < 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 0)$

Decreasing on: $(0, \infty)$

Step 9