Calculus Examples

$f (x) = 3 (5^{\sqrt{2 x + 7}})$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 7}$ as ${(2 x + 7)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}}]$

Step 1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}}$ with respect to $x$ is $3 \frac{d}{d x} [5^{{(2 x + 7)}^{\frac{1}{2}}}]$ .

$3 \frac{d}{d x} [5^{{(2 x + 7)}^{\frac{1}{2}}}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 5^{x}$ and $g (x) = {(2 x + 7)}^{\frac{1}{2}}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as ${(2 x + 7)}^{\frac{1}{2}}$ .

$3 (\frac{d}{d u_{1}} [5^{u_{1}}] \frac{d}{d x} [{(2 x + 7)}^{\frac{1}{2}}])$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $5$ .

$3 (5^{u_{1}} \ln (5) \frac{d}{d x} [{(2 x + 7)}^{\frac{1}{2}}])$

Step 2.3

Replace all occurrences of $u_{1}$ with ${(2 x + 7)}^{\frac{1}{2}}$ .

$3 (5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) \frac{d}{d x} [{(2 x + 7)}^{\frac{1}{2}}])$

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 $g (x) = 2 x + 7$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $2 x + 7$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d x} [2 x + 7])$

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

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d x} [2 x + 7])$

Step 3.3

Replace all occurrences of $u_{2}$ with $2 x + 7$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{1}{2} - 1} \frac{d}{d x} [2 x + 7])$

Step 4

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

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [2 x + 7])$

Step 5

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

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [2 x + 7])$

Step 6

Combine the numerators over the common denominator.

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [2 x + 7])$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{1 - 2}{2}} \frac{d}{d x} [2 x + 7])$

Step 7.2

Subtract $2$ from $1$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{\frac{- 1}{2}} \frac{d}{d x} [2 x + 7])$

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2} {(2 x + 7)}^{- \frac{1}{2}} \frac{d}{d x} [2 x + 7])$

Step 8.2

Combine $\frac{1}{2}$ and ${(2 x + 7)}^{- \frac{1}{2}}$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{{(2 x + 7)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [2 x + 7])$

Step 8.3

Move ${(2 x + 7)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{1}{2 {(2 x + 7)}^{\frac{1}{2}}} \frac{d}{d x} [2 x + 7])$

Step 8.4

Combine $3$ and $\frac{1}{2 {(2 x + 7)}^{\frac{1}{2}}}$ .

$5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) (\frac{3}{2 {(2 x + 7)}^{\frac{1}{2}}} \frac{d}{d x} [2 x + 7])$

Step 8.5

Combine $\frac{3}{2 {(2 x + 7)}^{\frac{1}{2}}}$ and $5^{{(2 x + 7)}^{\frac{1}{2}}}$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}}}{2 {(2 x + 7)}^{\frac{1}{2}}} \ln (5) \frac{d}{d x} [2 x + 7]$

Step 8.6

Combine $\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}}}{2 {(2 x + 7)}^{\frac{1}{2}}}$ and $\ln (5)$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} \frac{d}{d x} [2 x + 7]$

Step 9

By the Sum Rule, the derivative of $2 x + 7$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [7]$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} (\frac{d}{d x} [2 x] + \frac{d}{d x} [7])$

Step 10

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

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} (2 \frac{d}{d x} [x] + \frac{d}{d x} [7])$

Step 11

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

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} (2 \cdot 1 + \frac{d}{d x} [7])$

Step 12

Multiply $2$ by $1$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} (2 + \frac{d}{d x} [7])$

Step 13

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} (2 + 0)$

Step 14

Simplify terms.

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Step 14.1

Add $2$ and $0$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}} \cdot 2$

Step 14.2

Combine $\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}}$ and $2$ .

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5) \cdot 2}{2 {(2 x + 7)}^{\frac{1}{2}}}$

Step 14.3

Multiply $2$ by $3$ .

$\frac{6 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{2 {(2 x + 7)}^{\frac{1}{2}}}$

Step 14.4

Factor $2$ out of $6 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)$ .

$\frac{2 (3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5))}{2 {(2 x + 7)}^{\frac{1}{2}}}$

Step 15

Cancel the common factors.

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Step 15.1

Factor $2$ out of $2 {(2 x + 7)}^{\frac{1}{2}}$ .

$\frac{2 (3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5))}{2 ({(2 x + 7)}^{\frac{1}{2}})}$

Step 15.2

Cancel the common factor.

$\frac{2 (3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5))}{2 {(2 x + 7)}^{\frac{1}{2}}}$

Step 15.3

Rewrite the expression.

$\frac{3 \cdot 5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5)}{{(2 x + 7)}^{\frac{1}{2}}}$

Step 16

Simplify the numerator.

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Step 16.1

Simplify $3 \ln (5)$ by moving $3$ inside the logarithm.

$\frac{5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (5^{3})}{{(2 x + 7)}^{\frac{1}{2}}}$

Step 16.2

Raise $5$ to the power of $3$ .

$\frac{5^{{(2 x + 7)}^{\frac{1}{2}}} \ln (125)}{{(2 x + 7)}^{\frac{1}{2}}}$