Calculus Examples

$z = {(3 x + 1)}^{4} {(5 x - 2)}^{3}$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d z} [f (z) g (z)]$ is $f (z) \frac{d}{d z} [g (z)] + g (z) \frac{d}{d z} [f (z)]$ where $f (z) = {(3 x + 1)}^{4}$ and $g (z) = {(5 x - 2)}^{3}$ .

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

Step 2

Use the Binomial Theorem.

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

Step 3

Simplify each term.

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Step 3.1

Apply the product rule to $5 x$ .

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

Step 3.2

Raise $5$ to the power of $3$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} + 3 {(5 x)}^{2} \cdot - 2 + 3 (5 x) {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.3

Apply the product rule to $5 x$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} + 3 (5^{2} x^{2}) \cdot - 2 + 3 (5 x) {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.4

Raise $5$ to the power of $2$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} + 3 (25 x^{2}) \cdot - 2 + 3 (5 x) {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.5

Multiply $25$ by $3$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} + 75 x^{2} \cdot - 2 + 3 (5 x) {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.6

Multiply $- 2$ by $75$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} - 150 x^{2} + 3 (5 x) {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.7

Multiply $5$ by $3$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} - 150 x^{2} + 15 x {(- 2)}^{2} + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.8

Raise $- 2$ to the power of $2$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} - 150 x^{2} + 15 x \cdot 4 + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.9

Multiply $4$ by $15$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} - 150 x^{2} + 60 x + {(- 2)}^{3}] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 3.10

Raise $- 2$ to the power of $3$ .

${(3 x + 1)}^{4} \frac{d}{d z} [125 x^{3} - 150 x^{2} + 60 x - 8] + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x + 1)}^{4}]$

Step 4

Since $125 x^{3} - 150 x^{2} + 60 x - 8$ is constant with respect to $z$ , the derivative of $125 x^{3} - 150 x^{2} + 60 x - 8$ with respect to $z$ is $0$ .

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

Step 5

Use the Binomial Theorem.

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [{(3 x)}^{4} + 4 {(3 x)}^{3} \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6

Simplify each term.

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Step 6.1

Apply the product rule to $3 x$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [3^{4} x^{4} + 4 {(3 x)}^{3} \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.2

Raise $3$ to the power of $4$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 4 {(3 x)}^{3} \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.3

Apply the product rule to $3 x$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 4 (3^{3} x^{3}) \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.4

Raise $3$ to the power of $3$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 4 (27 x^{3}) \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.5

Multiply $27$ by $4$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} \cdot 1 + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.6

Multiply $108$ by $1$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 6 {(3 x)}^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.7

Apply the product rule to $3 x$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 6 (3^{2} x^{2}) \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.8

Raise $3$ to the power of $2$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 6 (9 x^{2}) \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.9

Multiply $9$ by $6$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} \cdot 1^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.10

One to any power is one.

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} \cdot 1 + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.11

Multiply $54$ by $1$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} + 4 (3 x) \cdot 1^{3} + 1^{4}]$

Step 6.12

Multiply $3$ by $4$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x \cdot 1^{3} + 1^{4}]$

Step 6.13

One to any power is one.

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x \cdot 1 + 1^{4}]$

Step 6.14

Multiply $12$ by $1$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x + 1^{4}]$

Step 6.15

One to any power is one.

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \frac{d}{d z} [81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x + 1]$

Step 7

Since $81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x + 1$ is constant with respect to $z$ , the derivative of $81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x + 1$ with respect to $z$ is $0$ .

${(3 x + 1)}^{4} \cdot 0 + {(5 x - 2)}^{3} \cdot 0$

Step 8

Combine terms.

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Step 8.1

Multiply ${(3 x + 1)}^{4}$ by $0$ .

$0 + {(5 x - 2)}^{3} \cdot 0$

Step 8.2

Multiply ${(5 x - 2)}^{3}$ by $0$ .

$0 + 0$

Step 8.3

Add $0$ and $0$ .

$0$