Calculus Examples

$g (x) = {(3 x - 1)}^{7} {(2 x + 1)}^{5}$

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 $f (x) = {(3 x - 1)}^{7}$ and $g (x) = {(2 x + 1)}^{5}$ .

${(3 x - 1)}^{7} \frac{d}{d x} [{(2 x + 1)}^{5}] + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = 2 x + 1$ .

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2 x + 1$ .

${(3 x - 1)}^{7} (\frac{d}{d u_{1}} [{u_{1}}^{5}] \frac{d}{d x} [2 x + 1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 2.2

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

${(3 x - 1)}^{7} (5 {u_{1}}^{4} \frac{d}{d x} [2 x + 1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 2.3

Replace all occurrences of $u_{1}$ with $2 x + 1$ .

${(3 x - 1)}^{7} (5 {(2 x + 1)}^{4} \frac{d}{d x} [2 x + 1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3

Differentiate.

Tap for more steps...

Step 3.1

Move $5$ to the left of ${(3 x - 1)}^{7}$ .

$5 \cdot {(3 x - 1)}^{7} {(2 x + 1)}^{4} \frac{d}{d x} [2 x + 1] + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} (\frac{d}{d x} [2 x] + \frac{d}{d x} [1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.3

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

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} (2 \frac{d}{d x} [x] + \frac{d}{d x} [1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.4

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

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} (2 \cdot 1 + \frac{d}{d x} [1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.5

Multiply $2$ by $1$ .

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} (2 + \frac{d}{d x} [1]) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} (2 + 0) + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.7

Simplify the expression.

Tap for more steps...

Step 3.7.1

Add $2$ and $0$ .

$5 {(3 x - 1)}^{7} {(2 x + 1)}^{4} \cdot 2 + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 3.7.2

Multiply $2$ by $5$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + {(2 x + 1)}^{5} \frac{d}{d x} [{(3 x - 1)}^{7}]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{7}$ and $g (x) = 3 x - 1$ .

Tap for more steps...

Step 4.1

To apply the Chain Rule, set $u_{2}$ as $3 x - 1$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + {(2 x + 1)}^{5} (\frac{d}{d u_{2}} [{u_{2}}^{7}] \frac{d}{d x} [3 x - 1])$

Step 4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 7$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + {(2 x + 1)}^{5} (7 {u_{2}}^{6} \frac{d}{d x} [3 x - 1])$

Step 4.3

Replace all occurrences of $u_{2}$ with $3 x - 1$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + {(2 x + 1)}^{5} (7 {(3 x - 1)}^{6} \frac{d}{d x} [3 x - 1])$

Step 5

Differentiate.

Tap for more steps...

Step 5.1

Move $7$ to the left of ${(2 x + 1)}^{5}$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 \cdot {(2 x + 1)}^{5} {(3 x - 1)}^{6} \frac{d}{d x} [3 x - 1]$

Step 5.2

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

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} (\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1])$

Step 5.3

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

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} (3 \frac{d}{d x} [x] + \frac{d}{d x} [- 1])$

Step 5.4

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

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} (3 \cdot 1 + \frac{d}{d x} [- 1])$

Step 5.5

Multiply $3$ by $1$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} (3 + \frac{d}{d x} [- 1])$

Step 5.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} (3 + 0)$

Step 5.7

Simplify with factoring out.

Tap for more steps...

Step 5.7.1

Add $3$ and $0$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 7 {(2 x + 1)}^{5} {(3 x - 1)}^{6} \cdot 3$

Step 5.7.2

Multiply $3$ by $7$ .

$10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 21 {(2 x + 1)}^{5} {(3 x - 1)}^{6}$

Step 5.7.3

Factor ${(3 x - 1)}^{6} {(2 x + 1)}^{4}$ out of $10 {(3 x - 1)}^{7} {(2 x + 1)}^{4} + 21 {(2 x + 1)}^{5} {(3 x - 1)}^{6}$ .

Tap for more steps...

Step 5.7.3.1

Factor ${(3 x - 1)}^{6} {(2 x + 1)}^{4}$ out of $10 {(3 x - 1)}^{7} {(2 x + 1)}^{4}$ .

${(3 x - 1)}^{6} {(2 x + 1)}^{4} (10 (3 x - 1)) + 21 {(2 x + 1)}^{5} {(3 x - 1)}^{6}$

Step 5.7.3.2

Factor ${(3 x - 1)}^{6} {(2 x + 1)}^{4}$ out of $21 {(2 x + 1)}^{5} {(3 x - 1)}^{6}$ .

${(3 x - 1)}^{6} {(2 x + 1)}^{4} (10 (3 x - 1)) + {(3 x - 1)}^{6} {(2 x + 1)}^{4} (21 (2 x + 1))$

Step 5.7.3.3

Factor ${(3 x - 1)}^{6} {(2 x + 1)}^{4}$ out of ${(3 x - 1)}^{6} {(2 x + 1)}^{4} (10 (3 x - 1)) + {(3 x - 1)}^{6} {(2 x + 1)}^{4} (21 (2 x + 1))$ .

${(3 x - 1)}^{6} {(2 x + 1)}^{4} (10 (3 x - 1) + 21 (2 x + 1))$