Calculus Examples

$f (x) = 3 - 6^{x^{1} \cdot 40}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $3 - 6^{x^{1} \cdot 40}$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- 6^{x^{1} \cdot 40}]$ .

$\frac{d}{d x} [3] + \frac{d}{d x} [- 6^{x^{1} \cdot 40}]$

Step 1.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 6^{x^{1} \cdot 40}]$

Step 2

Evaluate $\frac{d}{d x} [- 6^{x^{1} \cdot 40}]$ .

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Step 2.1

Simplify.

$0 + \frac{d}{d x} [- 6^{x \cdot 40}]$

Step 2.2

Move $40$ to the left of $x$ .

$0 + \frac{d}{d x} [- 6^{40 \cdot x}]$

Step 2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 6^{40 x}$ with respect to $x$ is $- \frac{d}{d x} [6^{40 x}]$ .

$0 - \frac{d}{d x} [6^{40 x}]$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 6^{x}$ and $g (x) = 40 x$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $40 x$ .

$0 - (\frac{d}{d u} [6^{u}] \frac{d}{d x} [40 x])$

Step 2.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $6$ .

$0 - (6^{40 x} \ln (6) \frac{d}{d x} [40 x])$

Step 2.5

Since $40$ is constant with respect to $x$ , the derivative of $40 x$ with respect to $x$ is $40 \frac{d}{d x} [x]$ .

$0 - (6^{40 x} \ln (6) (40 \frac{d}{d x} [x]))$

Step 2.6

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

$0 - (6^{40 x} \ln (6) (40 \cdot 1))$

Step 2.7

Multiply $40$ by $1$ .

$0 - (6^{40 x} \ln (6) \cdot 40)$

Step 2.8

Move $40$ to the left of $6^{40 x} \ln (6)$ .

$0 - (40 \cdot (6^{40 x} \ln (6)))$

Step 2.9

Multiply $40$ by $- 1$ .

$0 - 40 \cdot 6^{40 x} \ln (6)$

Step 3

Subtract $40 \cdot 6^{40 x} \ln (6)$ from $0$ .

$- 40 \cdot 6^{40 x} \ln (6)$