Calculus Examples

${(e^{8 x - 1})}^{5}$

Step 1

Multiply the exponents in ${(e^{8 x - 1})}^{5}$ .

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Step 1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{d}{d x} [e^{(8 x - 1) \cdot 5}]$

Step 1.2

Apply the distributive property.

$\frac{d}{d x} [e^{8 x \cdot 5 - 1 \cdot 5}]$

Step 1.3

Multiply $5$ by $8$ .

$\frac{d}{d x} [e^{40 x - 1 \cdot 5}]$

Step 1.4

Multiply $- 1$ by $5$ .

$\frac{d}{d x} [e^{40 x - 5}]$

Step 2

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 $g (x) = 40 x - 5$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $40 x - 5$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [40 x - 5]$

Step 2.2

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

$e^{u} \frac{d}{d x} [40 x - 5]$

Step 2.3

Replace all occurrences of $u$ with $40 x - 5$ .

$e^{40 x - 5} \frac{d}{d x} [40 x - 5]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $40 x - 5$ with respect to $x$ is $\frac{d}{d x} [40 x] + \frac{d}{d x} [- 5]$ .

$e^{40 x - 5} (\frac{d}{d x} [40 x] + \frac{d}{d x} [- 5])$

Step 3.2

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

$e^{40 x - 5} (40 \frac{d}{d x} [x] + \frac{d}{d x} [- 5])$

Step 3.3

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

$e^{40 x - 5} (40 \cdot 1 + \frac{d}{d x} [- 5])$

Step 3.4

Multiply $40$ by $1$ .

$e^{40 x - 5} (40 + \frac{d}{d x} [- 5])$

Step 3.5

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$e^{40 x - 5} (40 + 0)$

Step 3.6

Simplify the expression.

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Step 3.6.1

Add $40$ and $0$ .

$e^{40 x - 5} \cdot 40$

Step 3.6.2

Move $40$ to the left of $e^{40 x - 5}$ .

$40 e^{40 x - 5}$