Calculus Examples

$f (x) = (3 x - 8) \ln (2 x^{5} + 3)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 3 x - 8$ and $g (x) = \ln (2 x^{5} + 3)$ .

$(3 x - 8) \frac{d}{d x} [\ln (2 x^{5} + 3)] + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x^{5} + 3$ .

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u$ as $2 x^{5} + 3$ .

$(3 x - 8) (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x^{5} + 3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$(3 x - 8) (\frac{1}{u} \frac{d}{d x} [2 x^{5} + 3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 2.3

Replace all occurrences of $u$ with $2 x^{5} + 3$ .

$(3 x - 8) (\frac{1}{2 x^{5} + 3} \frac{d}{d x} [2 x^{5} + 3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3

Differentiate.

Tap for more steps...

Step 3.1

By the Sum Rule, the derivative of $2 x^{5} + 3$ with respect to $x$ is $\frac{d}{d x} [2 x^{5}] + \frac{d}{d x} [3]$ .

$(3 x - 8) \frac{1}{2 x^{5} + 3} (\frac{d}{d x} [2 x^{5}] + \frac{d}{d x} [3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{5}$ with respect to $x$ is $2 \frac{d}{d x} [x^{5}]$ .

$(3 x - 8) \frac{1}{2 x^{5} + 3} (2 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$(3 x - 8) \frac{1}{2 x^{5} + 3} (2 (5 x^{4}) + \frac{d}{d x} [3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.4

Multiply $5$ by $2$ .

$(3 x - 8) \frac{1}{2 x^{5} + 3} (10 x^{4} + \frac{d}{d x} [3]) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.5

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

$(3 x - 8) \frac{1}{2 x^{5} + 3} (10 x^{4} + 0) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.6

Combine fractions.

Tap for more steps...

Step 3.6.1

Add $10 x^{4}$ and $0$ .

$(3 x - 8) \frac{1}{2 x^{5} + 3} (10 x^{4}) + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.6.2

Combine $10$ and $\frac{1}{2 x^{5} + 3}$ .

$(3 x - 8) \frac{10}{2 x^{5} + 3} x^{4} + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.6.3

Combine $x^{4}$ and $\frac{10}{2 x^{5} + 3}$ .

$(3 x - 8) \frac{x^{4} \cdot 10}{2 x^{5} + 3} + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.6.4

Move $10$ to the left of $x^{4}$ .

$(3 x - 8) \frac{10 \cdot x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) \frac{d}{d x} [3 x - 8]$

Step 3.7

By the Sum Rule, the derivative of $3 x - 8$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 8]$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) (\frac{d}{d x} [3 x] + \frac{d}{d x} [- 8])$

Step 3.8

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) (3 \frac{d}{d x} [x] + \frac{d}{d x} [- 8])$

Step 3.9

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

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) (3 \cdot 1 + \frac{d}{d x} [- 8])$

Step 3.10

Multiply $3$ by $1$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) (3 + \frac{d}{d x} [- 8])$

Step 3.11

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) (3 + 0)$

Step 3.12

Simplify the expression.

Tap for more steps...

Step 3.12.1

Add $3$ and $0$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + \ln (2 x^{5} + 3) \cdot 3$

Step 3.12.2

Move $3$ to the left of $\ln (2 x^{5} + 3)$ .

$(3 x - 8) \frac{10 x^{4}}{2 x^{5} + 3} + 3 \ln (2 x^{5} + 3)$