Calculus Examples

$x = t {(2 t + 1)}^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (x) = \frac{d}{d t} (t {(2 t + 1)}^{2})$

Step 2

The derivative of $x$ with respect to $t$ is $x'$ .

$x'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Rewrite ${(2 t + 1)}^{2}$ as $(2 t + 1) (2 t + 1)$ .

$\frac{d}{d t} [t ((2 t + 1) (2 t + 1))]$

Step 3.2

Expand $(2 t + 1) (2 t + 1)$ using the FOIL Method.

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Step 3.2.1

Apply the distributive property.

$\frac{d}{d t} [t (2 t (2 t + 1) + 1 (2 t + 1))]$

Step 3.2.2

Apply the distributive property.

$\frac{d}{d t} [t (2 t (2 t) + 2 t \cdot 1 + 1 (2 t + 1))]$

Step 3.2.3

Apply the distributive property.

$\frac{d}{d t} [t (2 t (2 t) + 2 t \cdot 1 + 1 (2 t) + 1 \cdot 1)]$

Step 3.3

Simplify and combine like terms.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{d}{d t} [t (2 \cdot 2 t \cdot t + 2 t \cdot 1 + 1 (2 t) + 1 \cdot 1)]$

Step 3.3.1.2

Multiply $t$ by $t$ by adding the exponents.

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Step 3.3.1.2.1

Move $t$ .

$\frac{d}{d t} [t (2 \cdot 2 (t \cdot t) + 2 t \cdot 1 + 1 (2 t) + 1 \cdot 1)]$

Step 3.3.1.2.2

Multiply $t$ by $t$ .

$\frac{d}{d t} [t (2 \cdot 2 t^{2} + 2 t \cdot 1 + 1 (2 t) + 1 \cdot 1)]$

Step 3.3.1.3

Multiply $2$ by $2$ .

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

Step 3.3.1.4

Multiply $2$ by $1$ .

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

Step 3.3.1.5

Multiply $2 t$ by $1$ .

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

Step 3.3.1.6

Multiply $1$ by $1$ .

$\frac{d}{d t} [t (4 t^{2} + 2 t + 2 t + 1)]$

Step 3.3.2

Add $2 t$ and $2 t$ .

$\frac{d}{d t} [t (4 t^{2} + 4 t + 1)]$

Step 3.4

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

$t \frac{d}{d t} [4 t^{2} + 4 t + 1] + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5

Differentiate.

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Step 3.5.1

By the Sum Rule, the derivative of $4 t^{2} + 4 t + 1$ with respect to $t$ is $\frac{d}{d t} [4 t^{2}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [1]$ .

$t (\frac{d}{d t} [4 t^{2}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.2

Since $4$ is constant with respect to $t$ , the derivative of $4 t^{2}$ with respect to $t$ is $4 \frac{d}{d t} [t^{2}]$ .

$t (4 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [4 t] + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.3

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

$t (4 (2 t) + \frac{d}{d t} [4 t] + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.4

Multiply $2$ by $4$ .

$t (8 t + \frac{d}{d t} [4 t] + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.5

Since $4$ is constant with respect to $t$ , the derivative of $4 t$ with respect to $t$ is $4 \frac{d}{d t} [t]$ .

$t (8 t + 4 \frac{d}{d t} [t] + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.6

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

$t (8 t + 4 \cdot 1 + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.7

Multiply $4$ by $1$ .

$t (8 t + 4 + \frac{d}{d t} [1]) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.8

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

$t (8 t + 4 + 0) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.9

Add $8 t + 4$ and $0$ .

$t (8 t + 4) + (4 t^{2} + 4 t + 1) \frac{d}{d t} [t]$

Step 3.5.10

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

$t (8 t + 4) + (4 t^{2} + 4 t + 1) \cdot 1$

Step 3.5.11

Multiply $4 t^{2} + 4 t + 1$ by $1$ .

$t (8 t + 4) + 4 t^{2} + 4 t + 1$

Step 3.6

Simplify.

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Step 3.6.1

Apply the distributive property.

$t (8 t) + t \cdot 4 + 4 t^{2} + 4 t + 1$

Step 3.6.2

Combine terms.

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Step 3.6.2.1

Raise $t$ to the power of $1$ .

$8 (t^{1} t) + t \cdot 4 + 4 t^{2} + 4 t + 1$

Step 3.6.2.2

Raise $t$ to the power of $1$ .

$8 (t^{1} t^{1}) + t \cdot 4 + 4 t^{2} + 4 t + 1$

Step 3.6.2.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$8 t^{1 + 1} + t \cdot 4 + 4 t^{2} + 4 t + 1$

Step 3.6.2.4

Add $1$ and $1$ .

$8 t^{2} + t \cdot 4 + 4 t^{2} + 4 t + 1$

Step 3.6.2.5

Move $4$ to the left of $t$ .

$8 t^{2} + 4 \cdot t + 4 t^{2} + 4 t + 1$

Step 3.6.2.6

Add $8 t^{2}$ and $4 t^{2}$ .

$12 t^{2} + 4 t + 4 t + 1$

Step 3.6.2.7

Add $4 t$ and $4 t$ .

$12 t^{2} + 8 t + 1$

Step 4

Reform the equation by setting the left side equal to the right side.

$x' = 12 t^{2} + 8 t + 1$

Step 5

Replace $x'$ with $\frac{d x}{d t}$ .

$\frac{d x}{d t} = 12 t^{2} + 8 t + 1$