Calculus Examples

$f' (x) = \frac{6}{\sqrt{x}}$

Step 1

The function $f (x)$ can be found by evaluating the indefinite integral of the derivative $f' (x)$ .

$f (x) = \int f' (x) d x$

Step 2

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int \frac{1}{\sqrt{x}} d x$

Step 3

Apply basic rules of exponents.

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Step 3.1

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

$6 \int \frac{1}{x^{\frac{1}{2}}} d x$

Step 3.2

Move $x^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$6 \int {(x^{\frac{1}{2}})}^{- 1} d x$

Step 3.3

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

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Step 3.3.1

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

$6 \int x^{\frac{1}{2} \cdot - 1} d x$

Step 3.3.2

Combine $\frac{1}{2}$ and $- 1$ .

$6 \int x^{\frac{- 1}{2}} d x$

Step 3.3.3

Move the negative in front of the fraction.

$6 \int x^{- \frac{1}{2}} d x$

Step 4

By the Power Rule, the integral of $x^{- \frac{1}{2}}$ with respect to $x$ is $2 x^{\frac{1}{2}}$ .

$6 (2 x^{\frac{1}{2}} + C)$

Step 5

Simplify the answer.

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Step 5.1

Rewrite $6 (2 x^{\frac{1}{2}} + C)$ as $6 \cdot 2 x^{\frac{1}{2}} + C$ .

$6 \cdot 2 x^{\frac{1}{2}} + C$

Step 5.2

Multiply $6$ by $2$ .

$12 x^{\frac{1}{2}} + C$

Step 6

The function $f$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$f (x) = 12 x^{\frac{1}{2}} + C$