Calculus Examples

$e^{7 x}$

Write $e^{7 x}$ as a function.

$f (x) = e^{7 x}$

Find the first derivative.

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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 $g (x) = 7 x$ .

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To apply the Chain Rule, set $u$ as $7 x$ .

$f' (x) = \frac{d}{d u} (e^{u}) \frac{d}{d x} (7 x)$

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

$f' (x) = e^{u} \frac{d}{d x} (7 x)$

Replace all occurrences of $u$ with $7 x$ .

$f' (x) = e^{7 x} \frac{d}{d x} (7 x)$

Differentiate.

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Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$f' (x) = e^{7 x} (7 \frac{d}{d x} (x))$

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

$f' (x) = e^{7 x} (7 \cdot 1)$

Simplify the expression.

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Multiply $7$ by $1$ .

$f' (x) = e^{7 x} \cdot 7$

Move $7$ to the left of $e^{7 x}$ .

$f' (x) = 7 e^{7 x}$

The first derivative of $f (x)$ with respect to $x$ is $7 e^{7 x}$ .

$7 e^{7 x}$

Set the first derivative equal to $0$ then solve the equation $7 e^{7 x} = 0$ .

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Set the first derivative equal to $0$ .

$7 e^{7 x} = 0$

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

$\ln (7 e^{7 x}) = \ln (0)$

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

Undefined

There is no solution for $7 e^{7 x} = 0$

No solution

No points make the derivative $f' (x) = 7 e^{7 x}$ equal to $0$ or undefined. The interval to check if $f (x) = e^{7 x}$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = 7 e^{7 x}$ to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval $(- \infty, \infty)$ . If the result is positive, the graph is increasing on the interval $(- \infty, \infty)$ .

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

$f' (1) = 7 e^{7 (1)}$

Simplify the result.

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Multiply $7$ by $1$ .

$f' (1) = 7 e^{7}$

The final answer is $7 e^{7}$ .

$7 e^{7}$

The result of substituting $1$ into $f' (x) = 7 e^{7 x}$ is $7 e^{7}$ , which is positive, so the graph is increasing on the interval $(- \infty, \infty)$ .

Increasing on $(- \infty, \infty)$ since $7 e^{7 x} > 0$

Increasing over the interval $(- \infty, \infty)$ means that the function is always increasing.

Always Increasing