Calculus Examples

$\log_{5} (\sqrt{3 x + 5})$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 5}$ as ${(3 x + 5)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [\log_{5} ({(3 x + 5)}^{\frac{1}{2}})]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log_{5} (x)$ and $g (x) = {(3 x + 5)}^{\frac{1}{2}}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as ${(3 x + 5)}^{\frac{1}{2}}$ .

$\frac{d}{d u_{1}} [\log_{5} (u_{1})] \frac{d}{d x} [{(3 x + 5)}^{\frac{1}{2}}]$

Step 2.2

The derivative of $\log_{5} (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1} \ln (5)}$ .

$\frac{1}{u_{1} \ln (5)} \frac{d}{d x} [{(3 x + 5)}^{\frac{1}{2}}]$

Step 2.3

Replace all occurrences of $u_{1}$ with ${(3 x + 5)}^{\frac{1}{2}}$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} \frac{d}{d x} [{(3 x + 5)}^{\frac{1}{2}}]$

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 $g (x) = 3 x + 5$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $3 x + 5$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d x} [3 x + 5])$

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 $n = \frac{1}{2}$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d x} [3 x + 5])$

Step 3.3

Replace all occurrences of $u_{2}$ with $3 x + 5$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{1}{2} - 1} \frac{d}{d x} [3 x + 5])$

Step 4

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

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [3 x + 5])$

Step 5

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

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [3 x + 5])$

Step 6

Combine the numerators over the common denominator.

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [3 x + 5])$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{1 - 2}{2}} \frac{d}{d x} [3 x + 5])$

Step 7.2

Subtract $2$ from $1$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{\frac{- 1}{2}} \frac{d}{d x} [3 x + 5])$

Step 8

Move the negative in front of the fraction.

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2} {(3 x + 5)}^{- \frac{1}{2}} \frac{d}{d x} [3 x + 5])$

Step 9

Combine $\frac{1}{2}$ and ${(3 x + 5)}^{- \frac{1}{2}}$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{{(3 x + 5)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [3 x + 5])$

Step 10

Move ${(3 x + 5)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)} (\frac{1}{2 {(3 x + 5)}^{\frac{1}{2}}} \frac{d}{d x} [3 x + 5])$

Step 11

Multiply $\frac{1}{2 {(3 x + 5)}^{\frac{1}{2}}}$ by $\frac{1}{{(3 x + 5)}^{\frac{1}{2}} \ln (5)}$ .

$\frac{1}{2 {(3 x + 5)}^{\frac{1}{2}} ({(3 x + 5)}^{\frac{1}{2}} \ln (5))} \frac{d}{d x} [3 x + 5]$

Step 12

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

$\frac{1}{2 {(3 x + 5)}^{\frac{1}{2} + \frac{1}{2}} \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 13

Simplify the expression.

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Step 13.1

Combine the numerators over the common denominator.

$\frac{1}{2 {(3 x + 5)}^{\frac{1 + 1}{2}} \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 13.2

Add $1$ and $1$ .

$\frac{1}{2 {(3 x + 5)}^{\frac{2}{2}} \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 14

Cancel the common factor of $2$ .

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Step 14.1

Cancel the common factor.

$\frac{1}{2 {(3 x + 5)}^{\frac{2}{2}} \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 14.2

Rewrite the expression.

$\frac{1}{2 {(3 x + 5)}^{1} \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 15

Simplify.

$\frac{1}{2 (3 x + 5) \ln (5)} \frac{d}{d x} [3 x + 5]$

Step 16

By the Sum Rule, the derivative of $3 x + 5$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [5]$ .

$\frac{1}{2 (3 x + 5) \ln (5)} (\frac{d}{d x} [3 x] + \frac{d}{d x} [5])$

Step 17

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$\frac{1}{2 (3 x + 5) \ln (5)} (3 \frac{d}{d x} [x] + \frac{d}{d x} [5])$

Step 18

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

$\frac{1}{2 (3 x + 5) \ln (5)} (3 \cdot 1 + \frac{d}{d x} [5])$

Step 19

Multiply $3$ by $1$ .

$\frac{1}{2 (3 x + 5) \ln (5)} (3 + \frac{d}{d x} [5])$

Step 20

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{1}{2 (3 x + 5) \ln (5)} (3 + 0)$

Step 21

Combine fractions.

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Step 21.1

Add $3$ and $0$ .

$\frac{1}{2 (3 x + 5) \ln (5)} \cdot 3$

Step 21.2

Combine $\frac{1}{2 (3 x + 5) \ln (5)}$ and $3$ .

$\frac{3}{2 (3 x + 5) \ln (5)}$

Step 22

Simplify.

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Step 22.1

Apply the distributive property.

$\frac{3}{(2 (3 x) + 2 \cdot 5) \ln (5)}$

Step 22.2

Apply the distributive property.

$\frac{3}{2 (3 x) \ln (5) + 2 \cdot 5 \ln (5)}$

Step 22.3

Combine terms.

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Step 22.3.1

Multiply $3$ by $2$ .

$\frac{3}{6 x \ln (5) + 2 \cdot 5 \ln (5)}$

Step 22.3.2

Multiply $2$ by $5$ .

$\frac{3}{6 x \ln (5) + 10 \ln (5)}$