Calculus Examples

$x = 2 t^{2} + t$

Step 1

Differentiate both sides of the equation.

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

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

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

$\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [t]$

Step 3.2

Evaluate $\frac{d}{d t} [2 t^{2}]$ .

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Step 3.2.1

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

$2 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [t]$

Step 3.2.2

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

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

Step 3.2.3

Multiply $2$ by $2$ .

$4 t + \frac{d}{d t} [t]$

Step 3.3

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

$4 t + 1$

Step 4

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

$x' = 4 t + 1$

Step 5

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

$\frac{d x}{d t} = 4 t + 1$