Calculus Examples

$e^{1 - 20 x + 5 x^{2}}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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 $g (x) = 1 - 20 x + 5 x^{2}$ .

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Step 1.1.1.1

To apply the Chain Rule, set $u$ as $1 - 20 x + 5 x^{2}$ .

$f' (x) = \frac{d}{d u} (e^{u}) \frac{d}{d x} (1 - 20 x + 5 x^{2})$

Step 1.1.1.2

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

$f' (x) = e^{u} \frac{d}{d x} (1 - 20 x + 5 x^{2})$

Step 1.1.1.3

Replace all occurrences of $u$ with $1 - 20 x + 5 x^{2}$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} (1 - 20 x + 5 x^{2})$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 - 20 x + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}]$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} (1) + \frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (0 + \frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.3

Add $0$ and $\frac{d}{d x} [- 20 x]$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.4

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x$ with respect to $x$ is $- 20 \frac{d}{d x} [x]$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 \frac{d}{d x} x + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.5

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

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 \cdot 1 + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.6

Multiply $- 20$ by $1$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 + \frac{d}{d x} (5 x^{2}))$

Step 1.1.2.7

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

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 + 5 \frac{d}{d x} (x^{2}))$

Step 1.1.2.8

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

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 + 5 (2 x))$

Step 1.1.2.9

Multiply $2$ by $5$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 2

Set the first derivative equal to $0$ then solve the equation $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$

Step 2.2

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

$e^{1 - 20 x + 5 x^{2}} = 0$

$- 20 + 10 x = 0$

Step 2.3

Set $e^{1 - 20 x + 5 x^{2}}$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $e^{1 - 20 x + 5 x^{2}}$ equal to $0$ .

$e^{1 - 20 x + 5 x^{2}} = 0$

Step 2.3.2

Solve $e^{1 - 20 x + 5 x^{2}} = 0$ for $x$ .

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Step 2.3.2.1

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

$\ln (e^{1 - 20 x + 5 x^{2}}) = \ln (0)$

Step 2.3.2.2

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

Undefined

Step 2.3.2.3

There is no solution for $e^{1 - 20 x + 5 x^{2}} = 0$

No solution

Step 2.4

Set $- 20 + 10 x$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $- 20 + 10 x$ equal to $0$ .

$- 20 + 10 x = 0$

Step 2.4.2

Solve $- 20 + 10 x = 0$ for $x$ .

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Step 2.4.2.1

Add $20$ to both sides of the equation.

$10 x = 20$

Step 2.4.2.2

Divide each term in $10 x = 20$ by $10$ and simplify.

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Step 2.4.2.2.1

Divide each term in $10 x = 20$ by $10$ .

$\frac{10 x}{10} = \frac{20}{10}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $10$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{10 x}{10} = \frac{20}{10}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{20}{10}$

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Divide $20$ by $10$ .

$x = 2$

Step 2.5

The final solution is all the values that make $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$ true.

$x = 2$

Step 3

Find the values where the derivative is undefined.

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Step 3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 4

Evaluate $e^{1 - 20 x + 5 x^{2}}$ at each $x$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $x = 2$ .

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Step 4.1.1

Substitute $2$ for $x$ .

$e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 4.1.2

Simplify.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

Multiply $- 20$ by $2$ .

$e^{1 - 40 + 5 {(2)}^{2}}$

Step 4.1.2.1.2

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

$e^{1 - 40 + 5 \cdot 4}$

Step 4.1.2.1.3

Multiply $5$ by $4$ .

$e^{1 - 40 + 20}$

Step 4.1.2.2

Simplify by adding and subtracting.

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Step 4.1.2.2.1

Subtract $40$ from $1$ .

$e^{- 39 + 20}$

Step 4.1.2.2.2

Add $- 39$ and $20$ .

$e^{- 19}$

Step 4.1.2.3

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

$\frac{1}{e^{19}}$

Step 4.2

List all of the points.

$(2, \frac{1}{e^{19}})$

Step 5