Calculus Examples

$f (t) = \sqrt{8 t^{2}} - t$

Step 1

Find the first derivative.

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Step 1.1

By the Sum Rule, the derivative of $\sqrt{8 t^{2}} - t$ with respect to $t$ is $\frac{d}{d t} [\sqrt{8 t^{2}}] + \frac{d}{d t} [- t]$ .

$\frac{d}{d t} [\sqrt{8 t^{2}}] + \frac{d}{d t} [- t]$

Step 1.2

Evaluate $\frac{d}{d t} [\sqrt{8 t^{2}}]$ .

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Step 1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 t^{2}}$ as ${(8 t^{2})}^{\frac{1}{2}}$ .

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

Step 1.2.2

Factor $8$ out of $8 t^{2}$ .

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

Step 1.2.3

Apply the product rule to $8 (t^{2})$ .

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

Step 1.2.4

Multiply the exponents in ${(t^{2})}^{\frac{1}{2}}$ .

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Step 1.2.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 1.2.4.2

Cancel the common factor of $2$ .

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Step 1.2.4.2.1

Cancel the common factor.

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

Step 1.2.4.2.2

Rewrite the expression.

$\frac{d}{d t} [8^{\frac{1}{2}} t^{1}] + \frac{d}{d t} [- t]$

Step 1.2.5

Simplify.

$\frac{d}{d t} [8^{\frac{1}{2}} t] + \frac{d}{d t} [- t]$

Step 1.2.6

Since $8^{\frac{1}{2}}$ is constant with respect to $t$ , the derivative of $8^{\frac{1}{2}} t$ with respect to $t$ is $8^{\frac{1}{2}} \frac{d}{d t} [t]$ .

$8^{\frac{1}{2}} \frac{d}{d t} [t] + \frac{d}{d t} [- t]$

Step 1.2.7

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

$8^{\frac{1}{2}} \cdot 1 + \frac{d}{d t} [- t]$

Step 1.2.8

Multiply $8^{\frac{1}{2}}$ by $1$ .

$8^{\frac{1}{2}} + \frac{d}{d t} [- t]$

Step 1.3

Evaluate $\frac{d}{d t} [- t]$ .

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Step 1.3.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$8^{\frac{1}{2}} - \frac{d}{d t} [t]$

Step 1.3.2

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

$8^{\frac{1}{2}} - 1 \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

$f' (t) = 8^{\frac{1}{2}} - 1$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $8^{\frac{1}{2}} - 1$ with respect to $t$ is $\frac{d}{d t} [8^{\frac{1}{2}}] + \frac{d}{d t} [- 1]$ .

$\frac{d}{d t} [8^{\frac{1}{2}}] + \frac{d}{d t} [- 1]$

Step 2.2

Since $8^{\frac{1}{2}}$ is constant with respect to $t$ , the derivative of $8^{\frac{1}{2}}$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [- 1]$

Step 2.3

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

$0 + 0$

Step 2.4

Add $0$ and $0$ .

$f'' (t) = 0$

Step 3

The second derivative of $f (t)$ with respect to $t$ is $0$ .

$0$