Calculus Examples

$f (x) = 2 \sqrt{x}$

Step 1

Consider the limit definition of the derivative.

$f' (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at $x = x + h$ .

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Step 2.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = 2 \sqrt{x + h}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Remove parentheses.

$f (x + h) = 2 \sqrt{x + h}$

Step 2.1.2.2

The final answer is $2 \sqrt{x + h}$ .

$2 \sqrt{x + h}$

Step 2.2

Find the components of the definition.

$f (x + h) = 2 \sqrt{x + h}$

$f (x) = 2 \sqrt{x}$

$f (x + h) = 2 \sqrt{x + h}$

$f (x) = 2 \sqrt{x}$

Step 3

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{2 \sqrt{x + h} - (2 \sqrt{x})}{h}$

Step 4

Simplify the numerator.

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Step 4.1

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

$f' (x) = lim_{h \to 0} \frac{2 {(x + h)}^{\frac{1}{2}} - 1 \cdot (2 \sqrt{x})}{h}$

Step 4.2

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

$f' (x) = lim_{h \to 0} \frac{2 {(x + h)}^{\frac{1}{2}} - 1 \cdot (2 x^{\frac{1}{2}})}{h}$

Step 4.3

Multiply $- 1$ by $2$ .

$f' (x) = lim_{h \to 0} \frac{2 {(x + h)}^{\frac{1}{2}} - 2 x^{\frac{1}{2}}}{h}$

Step 4.4

Factor $2$ out of $2 {(x + h)}^{\frac{1}{2}} - 2 x^{\frac{1}{2}}$ .

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Step 4.4.1

Factor $2$ out of $2 {(x + h)}^{\frac{1}{2}}$ .

$f' (x) = lim_{h \to 0} \frac{2 {(x + h)}^{\frac{1}{2}} - 2 x^{\frac{1}{2}}}{h}$

Step 4.4.2

Factor $2$ out of $- 2 x^{\frac{1}{2}}$ .

$f' (x) = lim_{h \to 0} \frac{2 {(x + h)}^{\frac{1}{2}} + 2 (- x^{\frac{1}{2}})}{h}$

Step 4.4.3

Factor $2$ out of $2 {(x + h)}^{\frac{1}{2}} + 2 (- x^{\frac{1}{2}})$ .

$f' (x) = lim_{h \to 0} \frac{2 ({(x + h)}^{\frac{1}{2}} - x^{\frac{1}{2}})}{h}$

Step 5

Because there are no values to the left of $0$ in the domain of $\frac{2 ({(x + h)}^{\frac{1}{2}} - x^{\frac{1}{2}})}{h}$ , the limit does not exist.

$Does not exist$

Step 6