Calculus Examples

$f (x) = \log_{3} (\sqrt{x - 1})$

Step 1

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

$\frac{d}{d x} [\log_{3} ({(x - 1)}^{\frac{1}{2}})]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log_{3} (x)$ and $g (x) = {(x - 1)}^{\frac{1}{2}}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as ${(x - 1)}^{\frac{1}{2}}$ .

$\frac{d}{d u_{1}} [\log_{3} (u_{1})] \frac{d}{d x} [{(x - 1)}^{\frac{1}{2}}]$

Step 2.2

The derivative of $\log_{3} (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1} \ln (3)}$ .

$\frac{1}{u_{1} \ln (3)} \frac{d}{d x} [{(x - 1)}^{\frac{1}{2}}]$

Step 2.3

Replace all occurrences of $u_{1}$ with ${(x - 1)}^{\frac{1}{2}}$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} \frac{d}{d x} [{(x - 1)}^{\frac{1}{2}}]$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x - 1$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $x - 1$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d x} [x - 1])$

Step 3.2

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

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d x} [x - 1])$

Step 3.3

Replace all occurrences of $u_{2}$ with $x - 1$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{1}{2} - 1} \frac{d}{d x} [x - 1])$

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x - 1])$

Step 5

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

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x - 1])$

Step 6

Combine the numerators over the common denominator.

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x - 1])$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{1 - 2}{2}} \frac{d}{d x} [x - 1])$

Step 7.2

Subtract $2$ from $1$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{\frac{- 1}{2}} \frac{d}{d x} [x - 1])$

Step 8

Move the negative in front of the fraction.

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2} {(x - 1)}^{- \frac{1}{2}} \frac{d}{d x} [x - 1])$

Step 9

Combine $\frac{1}{2}$ and ${(x - 1)}^{- \frac{1}{2}}$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{{(x - 1)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x - 1])$

Step 10

Move ${(x - 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)} (\frac{1}{2 {(x - 1)}^{\frac{1}{2}}} \frac{d}{d x} [x - 1])$

Step 11

Multiply $\frac{1}{2 {(x - 1)}^{\frac{1}{2}}}$ by $\frac{1}{{(x - 1)}^{\frac{1}{2}} \ln (3)}$ .

$\frac{1}{2 {(x - 1)}^{\frac{1}{2}} ({(x - 1)}^{\frac{1}{2}} \ln (3))} \frac{d}{d x} [x - 1]$

Step 12

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

$\frac{1}{2 {(x - 1)}^{\frac{1}{2} + \frac{1}{2}} \ln (3)} \frac{d}{d x} [x - 1]$

Step 13

Simplify the expression.

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Step 13.1

Combine the numerators over the common denominator.

$\frac{1}{2 {(x - 1)}^{\frac{1 + 1}{2}} \ln (3)} \frac{d}{d x} [x - 1]$

Step 13.2

Add $1$ and $1$ .

$\frac{1}{2 {(x - 1)}^{\frac{2}{2}} \ln (3)} \frac{d}{d x} [x - 1]$

Step 14

Cancel the common factor of $2$ .

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Step 14.1

Cancel the common factor.

$\frac{1}{2 {(x - 1)}^{\frac{2}{2}} \ln (3)} \frac{d}{d x} [x - 1]$

Step 14.2

Rewrite the expression.

$\frac{1}{2 {(x - 1)}^{1} \ln (3)} \frac{d}{d x} [x - 1]$

Step 15

Simplify.

$\frac{1}{2 (x - 1) \ln (3)} \frac{d}{d x} [x - 1]$

Step 16

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

$\frac{1}{2 (x - 1) \ln (3)} (\frac{d}{d x} [x] + \frac{d}{d x} [- 1])$

Step 17

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

$\frac{1}{2 (x - 1) \ln (3)} (1 + \frac{d}{d x} [- 1])$

Step 18

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

$\frac{1}{2 (x - 1) \ln (3)} (1 + 0)$

Step 19

Simplify the expression.

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Step 19.1

Add $1$ and $0$ .

$\frac{1}{2 (x - 1) \ln (3)} \cdot 1$

Step 19.2

Multiply $\frac{1}{2 (x - 1) \ln (3)}$ by $1$ .

$\frac{1}{2 (x - 1) \ln (3)}$

Step 20

Simplify.

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Step 20.1

Apply the distributive property.

$\frac{1}{(2 x + 2 \cdot - 1) \ln (3)}$

Step 20.2

Apply the distributive property.

$\frac{1}{2 x \ln (3) + 2 \cdot - 1 \ln (3)}$

Step 20.3

Multiply $2$ by $- 1$ .

$\frac{1}{2 x \ln (3) - 2 \ln (3)}$