Calculus Examples

$f (x) = \sqrt{3 x - 7}$

Step 1

Find the first derivative.

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x - 7}$ as ${(3 x - 7)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} ({(3 x - 7)}^{\frac{1}{2}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 3 x - 7$ .

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To apply the Chain Rule, set $u$ as $3 x - 7$ .

$f' (x) = \frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (3 x - 7)$

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

$f' (x) = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (3 x - 7)$

Replace all occurrences of $u$ with $3 x - 7$ .

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{1}{2} - 1} \frac{d}{d x} (3 x - 7)$

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

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (3 x - 7)$

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

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (3 x - 7)$

Combine the numerators over the common denominator.

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (3 x - 7)$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{1 - 2}{2}} \frac{d}{d x} (3 x - 7)$

Subtract $2$ from $1$ .

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{\frac{- 1}{2}} \frac{d}{d x} (3 x - 7)$

Combine fractions.

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Move the negative in front of the fraction.

$f' (x) = \frac{1}{2} \cdot {(3 x - 7)}^{- \frac{1}{2}} \frac{d}{d x} (3 x - 7)$

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

$f' (x) = \frac{{(3 x - 7)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (3 x - 7)$

Move ${(3 x - 7)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (3 x - 7)$

By the Sum Rule, the derivative of $3 x - 7$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 7]$ .

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (3 x) + \frac{d}{d x} (- 7))$

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

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot (3 \frac{d}{d x} (x) + \frac{d}{d x} (- 7))$

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

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot (3 \cdot 1 + \frac{d}{d x} (- 7))$

Multiply $3$ by $1$ .

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot (3 + \frac{d}{d x} (- 7))$

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

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot (3 + 0)$

Combine fractions.

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Add $3$ and $0$ .

$f' (x) = \frac{1}{2 {(3 x - 7)}^{\frac{1}{2}}} \cdot 3$

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

$f' (x) = \frac{3}{2 {(3 x - 7)}^{\frac{1}{2}}}$

Step 2

Find the second derivative.

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Differentiate using the Constant Multiple Rule.

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

$f'' (x) = \frac{3}{2} \cdot \frac{d}{d x} (\frac{1}{{(3 x - 7)}^{\frac{1}{2}}})$

Apply basic rules of exponents.

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Rewrite $\frac{1}{{(3 x - 7)}^{\frac{1}{2}}}$ as ${({(3 x - 7)}^{\frac{1}{2}})}^{- 1}$ .

$f'' (x) = \frac{3}{2} \cdot \frac{d}{d x} ({({(3 x - 7)}^{\frac{1}{2}})}^{- 1})$

Multiply the exponents in ${({(3 x - 7)}^{\frac{1}{2}})}^{- 1}$ .

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

$f'' (x) = \frac{3}{2} \cdot \frac{d}{d x} ({(3 x - 7)}^{\frac{1}{2} \cdot - 1})$

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

$f'' (x) = \frac{3}{2} \cdot \frac{d}{d x} ({(3 x - 7)}^{\frac{- 1}{2}})$

Move the negative in front of the fraction.

$f'' (x) = \frac{3}{2} \cdot \frac{d}{d x} ({(3 x - 7)}^{- \frac{1}{2}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{2}}$ and $g (x) = 3 x - 7$ .

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To apply the Chain Rule, set $u$ as $3 x - 7$ .

$f'' (x) = \frac{3}{2} \cdot (\frac{d}{d u} (u^{- \frac{1}{2}}) \frac{d}{d x} (3 x - 7))$

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

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} u^{- \frac{1}{2} - 1} \frac{d}{d x} (3 x - 7))$

Replace all occurrences of $u$ with $3 x - 7$ .

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{- \frac{1}{2} - 1} \frac{d}{d x} (3 x - 7))$

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

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{- \frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (3 x - 7))$

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

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{- \frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (3 x - 7))$

Combine the numerators over the common denominator.

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{\frac{- 1 - 1 \cdot 2}{2}} \frac{d}{d x} (3 x - 7))$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{\frac{- 1 - 2}{2}} \frac{d}{d x} (3 x - 7))$

Subtract $2$ from $- 1$ .

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{\frac{- 3}{2}} \frac{d}{d x} (3 x - 7))$

Combine fractions.

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Move the negative in front of the fraction.

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2} \cdot {(3 x - 7)}^{- \frac{3}{2}} \frac{d}{d x} (3 x - 7))$

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

$f'' (x) = \frac{3}{2} \cdot (- \frac{{(3 x - 7)}^{- \frac{3}{2}}}{2} \cdot \frac{d}{d x} (3 x - 7))$

Move ${(3 x - 7)}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{3}{2} \cdot (- \frac{1}{2 {(3 x - 7)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (3 x - 7))$

Multiply $\frac{3}{2}$ by $\frac{1}{2 {(3 x - 7)}^{\frac{3}{2}}}$ .

$f'' (x) = - \frac{3}{2 (2 {(3 x - 7)}^{\frac{3}{2}})} \cdot \frac{d}{d x} (3 x - 7)$

Multiply $2$ by $2$ .

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (3 x - 7)$

By the Sum Rule, the derivative of $3 x - 7$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 7]$ .

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot (\frac{d}{d x} (3 x) + \frac{d}{d x} (- 7))$

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

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot (3 \frac{d}{d x} (x) + \frac{d}{d x} (- 7))$

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

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot (3 \cdot 1 + \frac{d}{d x} (- 7))$

Multiply $3$ by $1$ .

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot (3 + \frac{d}{d x} (- 7))$

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

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot (3 + 0)$

Combine fractions.

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Add $3$ and $0$ .

$f'' (x) = - \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}} \cdot 3$

Multiply $3$ by $- 1$ .

$f'' (x) = - 3 \frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}}$

Combine $- 3$ and $\frac{3}{4 {(3 x - 7)}^{\frac{3}{2}}}$ .

$f'' (x) = \frac{- 3 \cdot 3}{4 {(3 x - 7)}^{\frac{3}{2}}}$

Simplify the expression.

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Multiply $- 3$ by $3$ .

$f'' (x) = \frac{- 9}{4 {(3 x - 7)}^{\frac{3}{2}}}$

Move the negative in front of the fraction.

$f'' (x) = - \frac{9}{4 {(3 x - 7)}^{\frac{3}{2}}}$

Step 3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{9}{4 {(3 x - 7)}^{\frac{3}{2}}}$ .

$- \frac{9}{4 {(3 x - 7)}^{\frac{3}{2}}}$