Calculus Examples

$f (x) = 3^{- x}$

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 $f (x) = 3^{x}$ and $g (x) = - x$ .

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Step 1.1.1.1

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

$f' (x) = \frac{d}{d u} (3^{u}) \frac{d}{d x} (- x)$

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 $a$ = $3$ .

$f' (x) = 3^{u} \ln (3) \frac{d}{d x} (- x)$

Step 1.1.1.3

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

$f' (x) = 3^{- x} \ln (3) \frac{d}{d x} (- x)$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f' (x) = 3^{- x} \ln (3) (- \frac{d}{d x} x)$

Step 1.1.2.2

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

$f' (x) = 3^{- x} \ln (3) (- 1 \cdot 1)$

Step 1.1.2.3

Simplify the expression.

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Step 1.1.2.3.1

Multiply $- 1$ by $1$ .

$f' (x) = 3^{- x} \ln (3) \cdot - 1$

Step 1.1.2.3.2

Move $- 1$ to the left of $3^{- x} \ln (3)$ .

$f' (x) = - 1 \cdot (3^{- x} \ln (3))$

Step 1.1.2.3.3

Rewrite $- 1 \cdot 3^{- x}$ as $- 3^{- x}$ .

$f' (x) = - 3^{- x} \ln (3)$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 3^{- x} \ln (3)$ .

$- 3^{- x} \ln (3)$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 3^{- x} \ln (3) = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 3^{- x} \ln (3) = 0$

Step 2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 3

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

No critical points found

Step 4

No points make the derivative $f' (x) = - 3^{- x} \ln (3)$ equal to $0$ or undefined. The interval to check if $f (x) = 3^{- x}$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Step 5

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = - 3^{- x} \ln (3)$ 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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Step 5.1

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

$f' (1) = - 3^{- (1)} \ln (3)$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $- 1$ by $1$ .

$f' (1) = - 3^{- 1} \ln (3)$

Step 5.2.2

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

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

Step 5.2.3

Multiply $- \frac{1}{3} \ln (3)$ .

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Step 5.2.3.1

Reorder $\ln (3)$ and $\frac{1}{3}$ .

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

Step 5.2.3.2

Simplify $\frac{1}{3} \ln (3)$ by moving $\frac{1}{3}$ inside the logarithm.

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

Step 5.2.4

The final answer is $- \ln (3^{\frac{1}{3}})$ .

$- \ln (3^{\frac{1}{3}})$

Step 6

The result of substituting $1$ into $f' (x) = - 3^{- x} \ln (3)$ is $- \ln (3^{\frac{1}{3}})$ , which is negative, so the graph is decreasing on the interval $(- \infty, \infty)$ .

Decreasing on $(- \infty, \infty)$

Step 7

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

Always Decreasing

Step 8