Calculus Examples

Let $h (x) = e^{2 x - 6} - e$

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

By the Sum Rule, the derivative of $e^{2 x - 6} - e$ with respect to $x$ is $\frac{d}{d x} [e^{2 x - 6}] + \frac{d}{d x} [- e]$ .

$\frac{d}{d x} [e^{2 x - 6}] + \frac{d}{d x} [- e]$

Step 1.1.2

Evaluate $\frac{d}{d x} [e^{2 x - 6}]$ .

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Step 1.1.2.1

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 $g (x) = 2 x - 6$ .

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Step 1.1.2.1.1

To apply the Chain Rule, set $u$ as $2 x - 6$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x - 6] + \frac{d}{d x} [- e]$

Step 1.1.2.1.2

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

$e^{u} \frac{d}{d x} [2 x - 6] + \frac{d}{d x} [- e]$

Step 1.1.2.1.3

Replace all occurrences of $u$ with $2 x - 6$ .

$e^{2 x - 6} \frac{d}{d x} [2 x - 6] + \frac{d}{d x} [- e]$

Step 1.1.2.2

By the Sum Rule, the derivative of $2 x - 6$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6]$ .

$e^{2 x - 6} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6]) + \frac{d}{d x} [- e]$

Step 1.1.2.3

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

$e^{2 x - 6} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 6]) + \frac{d}{d x} [- e]$

Step 1.1.2.4

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

$e^{2 x - 6} (2 \cdot 1 + \frac{d}{d x} [- 6]) + \frac{d}{d x} [- e]$

Step 1.1.2.5

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$e^{2 x - 6} (2 \cdot 1 + 0) + \frac{d}{d x} [- e]$

Step 1.1.2.6

Multiply $2$ by $1$ .

$e^{2 x - 6} (2 + 0) + \frac{d}{d x} [- e]$

Step 1.1.2.7

Add $2$ and $0$ .

$e^{2 x - 6} \cdot 2 + \frac{d}{d x} [- e]$

Step 1.1.2.8

Move $2$ to the left of $e^{2 x - 6}$ .

$2 e^{2 x - 6} + \frac{d}{d x} [- e]$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

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

$2 e^{2 x - 6} + 0$

Step 1.1.3.2

Add $2 e^{2 x - 6}$ and $0$ .

$f' (x) = 2 e^{2 x - 6}$

Step 1.2

The first derivative of $h (x)$ with respect to $x$ is $2 e^{2 x - 6}$ .

$2 e^{2 x - 6}$

Step 2

Set the first derivative equal to $0$ then solve the equation $2 e^{2 x - 6} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$2 e^{2 x - 6} = 0$

Step 2.2

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

$\ln (2 e^{2 x - 6}) = \ln (0)$

Step 2.3

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

Undefined

Step 2.4

There is no solution for $2 e^{2 x - 6} = 0$

No solution

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

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found