Calculus Examples

$y = x {(2 x + 3)}^{5}$

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

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 $g (x) = 2 x + 3$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $2 x + 3$ .

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

Step 1.2.2

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

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

Step 1.2.3

Replace all occurrences of $u$ with $2 x + 3$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Multiply $2$ by $1$ .

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

Step 1.3.5

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

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

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Add $2$ and $0$ .

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

Step 1.3.6.2

Multiply $2$ by $5$ .

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

Step 1.3.7

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

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

Step 1.3.8

Multiply ${(2 x + 3)}^{5}$ by $1$ .

$x (10 {(2 x + 3)}^{4}) + {(2 x + 3)}^{5}$

Step 1.4

Simplify.

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Step 1.4.1

Factor ${(2 x + 3)}^{4}$ out of $x (10 {(2 x + 3)}^{4}) + {(2 x + 3)}^{5}$ .

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Step 1.4.1.1

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

${(2 x + 3)}^{4} (x \cdot (10)) + {(2 x + 3)}^{5}$

Step 1.4.1.2

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

${(2 x + 3)}^{4} (x \cdot (10)) + {(2 x + 3)}^{4} (2 x + 3)$

Step 1.4.1.3

Factor ${(2 x + 3)}^{4}$ out of ${(2 x + 3)}^{4} (x \cdot (10)) + {(2 x + 3)}^{4} (2 x + 3)$ .

${(2 x + 3)}^{4} (x \cdot (10) + 2 x + 3)$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Move $10$ to the left of $x$ .

${(2 x + 3)}^{4} (10 \cdot x + 2 x + 3)$

Step 1.4.2.2

Add $10 x$ and $2 x$ .

$f' (x) = {(2 x + 3)}^{4} (12 x + 3)$

Step 2

Find the second derivative.

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Step 2.1

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 $f (x) = {(2 x + 3)}^{4}$ and $g (x) = 12 x + 3$ .

${(2 x + 3)}^{4} \frac{d}{d x} [12 x + 3] + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $12 x + 3$ with respect to $x$ is $\frac{d}{d x} [12 x] + \frac{d}{d x} [3]$ .

${(2 x + 3)}^{4} (\frac{d}{d x} [12 x] + \frac{d}{d x} [3]) + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.2

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

${(2 x + 3)}^{4} (12 \frac{d}{d x} [x] + \frac{d}{d x} [3]) + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.3

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

${(2 x + 3)}^{4} (12 \cdot 1 + \frac{d}{d x} [3]) + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.4

Multiply $12$ by $1$ .

${(2 x + 3)}^{4} (12 + \frac{d}{d x} [3]) + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.5

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

${(2 x + 3)}^{4} (12 + 0) + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.6

Simplify the expression.

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Step 2.2.6.1

Add $12$ and $0$ .

${(2 x + 3)}^{4} \cdot 12 + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.2.6.2

Move $12$ to the left of ${(2 x + 3)}^{4}$ .

$12 \cdot {(2 x + 3)}^{4} + (12 x + 3) \frac{d}{d x} [{(2 x + 3)}^{4}]$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = 2 x + 3$ .

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Step 2.3.1

To apply the Chain Rule, set $u$ as $2 x + 3$ .

$12 {(2 x + 3)}^{4} + (12 x + 3) (\frac{d}{d u} [u^{4}] \frac{d}{d x} [2 x + 3])$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$12 {(2 x + 3)}^{4} + (12 x + 3) (4 u^{3} \frac{d}{d x} [2 x + 3])$

Step 2.3.3

Replace all occurrences of $u$ with $2 x + 3$ .

$12 {(2 x + 3)}^{4} + (12 x + 3) (4 {(2 x + 3)}^{3} \frac{d}{d x} [2 x + 3])$

Step 2.4

Differentiate.

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Step 2.4.1

Move $4$ to the left of $12 x + 3$ .

$12 {(2 x + 3)}^{4} + 4 \cdot (12 x + 3) {(2 x + 3)}^{3} \frac{d}{d x} [2 x + 3]$

Step 2.4.2

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

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} (\frac{d}{d x} [2 x] + \frac{d}{d x} [3])$

Step 2.4.3

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

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} (2 \frac{d}{d x} [x] + \frac{d}{d x} [3])$

Step 2.4.4

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

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} (2 \cdot 1 + \frac{d}{d x} [3])$

Step 2.4.5

Multiply $2$ by $1$ .

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} (2 + \frac{d}{d x} [3])$

Step 2.4.6

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

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} (2 + 0)$

Step 2.4.7

Simplify the expression.

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Step 2.4.7.1

Add $2$ and $0$ .

$12 {(2 x + 3)}^{4} + 4 (12 x + 3) {(2 x + 3)}^{3} \cdot 2$

Step 2.4.7.2

Multiply $2$ by $4$ .

$12 {(2 x + 3)}^{4} + 8 (12 x + 3) {(2 x + 3)}^{3}$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$12 {(2 x + 3)}^{4} + (8 (12 x) + 8 \cdot 3) {(2 x + 3)}^{3}$

Step 2.5.2

Multiply $12$ by $8$ .

$12 {(2 x + 3)}^{4} + (96 x + 8 \cdot 3) {(2 x + 3)}^{3}$

Step 2.5.3

Multiply $8$ by $3$ .

$12 {(2 x + 3)}^{4} + (96 x + 24) {(2 x + 3)}^{3}$

Step 2.5.4

Factor ${(2 x + 3)}^{3}$ out of $12 {(2 x + 3)}^{4} + (96 x + 24) {(2 x + 3)}^{3}$ .

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Step 2.5.4.1

Factor ${(2 x + 3)}^{3}$ out of $12 {(2 x + 3)}^{4}$ .

${(2 x + 3)}^{3} (12 (2 x + 3)) + (96 x + 24) {(2 x + 3)}^{3}$

Step 2.5.4.2

Factor ${(2 x + 3)}^{3}$ out of $(96 x + 24) {(2 x + 3)}^{3}$ .

${(2 x + 3)}^{3} (12 (2 x + 3)) + {(2 x + 3)}^{3} (96 x + 24)$

Step 2.5.4.3

Factor ${(2 x + 3)}^{3}$ out of ${(2 x + 3)}^{3} (12 (2 x + 3)) + {(2 x + 3)}^{3} (96 x + 24)$ .

$f'' (x) = {(2 x + 3)}^{3} (12 (2 x + 3) + 96 x + 24)$