Calculus Examples

$8^{x^{2} - x}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 8^{x}$ and $g (x) = x^{2} - x$ .

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Step 1.1

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

$\frac{d}{d u} [8^{u}] \frac{d}{d x} [x^{2} - x]$

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $8$ .

$8^{u} \ln (8) \frac{d}{d x} [x^{2} - x]$

Step 1.3

Replace all occurrences of $u$ with $x^{2} - x$ .

$8^{x^{2} - x} \ln (8) \frac{d}{d x} [x^{2} - x]$

Step 2

Differentiate.

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Step 2.1

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

$8^{x^{2} - x} \ln (8) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x])$

Step 2.2

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

$8^{x^{2} - x} \ln (8) (2 x + \frac{d}{d x} [- x])$

Step 2.3

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

$8^{x^{2} - x} \ln (8) (2 x - \frac{d}{d x} [x])$

Step 2.4

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

$8^{x^{2} - x} \ln (8) (2 x - 1 \cdot 1)$

Step 2.5

Simplify the expression.

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Step 2.5.1

Multiply $- 1$ by $1$ .

$8^{x^{2} - x} \ln (8) (2 x - 1)$

Step 2.5.2

Reorder the factors of $8^{x^{2} - x} \ln (8) (2 x - 1)$ .

$8^{x^{2} - x} (2 x - 1) \ln (8)$