Calculus Examples

$h (x) = \frac{1}{{(2 x - 3)}^{0.5}} + {(2 x - 3)}^{0.5}$

Step 1

By the Sum Rule, the derivative of $\frac{1}{{(2 x - 3)}^{0.5}} + {(2 x - 3)}^{0.5}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{{(2 x - 3)}^{0.5}}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$ .

$\frac{d}{d x} [\frac{1}{{(2 x - 3)}^{0.5}}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2

Evaluate $\frac{d}{d x} [\frac{1}{{(2 x - 3)}^{0.5}}]$ .

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Step 2.1

Rewrite $\frac{1}{{(2 x - 3)}^{0.5}}$ as ${({(2 x - 3)}^{0.5})}^{- 1}$ .

$\frac{d}{d x} [{({(2 x - 3)}^{0.5})}^{- 1}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = {(2 x - 3)}^{0.5}$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as ${(2 x - 3)}^{0.5}$ .

$\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [{(2 x - 3)}^{0.5}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.2.2

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

$- {u_{1}}^{- 2} \frac{d}{d x} [{(2 x - 3)}^{0.5}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with ${(2 x - 3)}^{0.5}$ .

$- {({(2 x - 3)}^{0.5})}^{- 2} \frac{d}{d x} [{(2 x - 3)}^{0.5}] + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{0.5}$ and $g (x) = 2 x - 3$ .

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Step 2.3.1

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

$- {({(2 x - 3)}^{0.5})}^{- 2} (\frac{d}{d u_{2}} [{u_{2}}^{0.5}] \frac{d}{d x} [2 x - 3]) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 0.5$ .

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {u_{2}}^{- 0.5} \frac{d}{d x} [2 x - 3]) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.3.3

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

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {(2 x - 3)}^{- 0.5} \frac{d}{d x} [2 x - 3]) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.4

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {(2 x - 3)}^{- 0.5} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3])) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.5

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

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {(2 x - 3)}^{- 0.5} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 3])) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.6

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

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + \frac{d}{d x} [- 3])) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.7

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$- {({(2 x - 3)}^{0.5})}^{- 2} (0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + 0)) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.8

Multiply the exponents in ${({(2 x - 3)}^{0.5})}^{- 2}$ .

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Step 2.8.1

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

$- {(2 x - 3)}^{0.5 \cdot - 2} (0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + 0)) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.8.2

Multiply $0.5$ by $- 2$ .

$- {(2 x - 3)}^{- 1} (0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + 0)) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.9

Multiply $2$ by $1$ .

$- {(2 x - 3)}^{- 1} (0.5 {(2 x - 3)}^{- 0.5} (2 + 0)) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.10

Add $2$ and $0$ .

$- {(2 x - 3)}^{- 1} (0.5 {(2 x - 3)}^{- 0.5} \cdot 2) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.11

Multiply $2$ by $0.5$ .

$- {(2 x - 3)}^{- 1} (1 {(2 x - 3)}^{- 0.5}) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.12

Multiply ${(2 x - 3)}^{- 0.5}$ by $1$ .

$- {(2 x - 3)}^{- 1} {(2 x - 3)}^{- 0.5} + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.13

Multiply ${(2 x - 3)}^{- 1}$ by ${(2 x - 3)}^{- 0.5}$ by adding the exponents.

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Step 2.13.1

Move ${(2 x - 3)}^{- 0.5}$ .

$- ({(2 x - 3)}^{- 0.5} {(2 x - 3)}^{- 1}) + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.13.2

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

$- {(2 x - 3)}^{- 0.5 - 1} + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 2.13.3

Subtract $1$ from $- 0.5$ .

$- {(2 x - 3)}^{- 1.5} + \frac{d}{d x} [{(2 x - 3)}^{0.5}]$

Step 3

Evaluate $\frac{d}{d x} [{(2 x - 3)}^{0.5}]$ .

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{0.5}$ and $g (x) = 2 x - 3$ .

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Step 3.1.1

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

$- {(2 x - 3)}^{- 1.5} + \frac{d}{d u_{3}} [{u_{3}}^{0.5}] \frac{d}{d x} [2 x - 3]$

Step 3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 0.5$ .

$- {(2 x - 3)}^{- 1.5} + 0.5 {u_{3}}^{- 0.5} \frac{d}{d x} [2 x - 3]$

Step 3.1.3

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

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} \frac{d}{d x} [2 x - 3]$

Step 3.2

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3])$

Step 3.3

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

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 3])$

Step 3.4

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

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + \frac{d}{d x} [- 3])$

Step 3.5

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} (2 \cdot 1 + 0)$

Step 3.6

Multiply $2$ by $1$ .

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} (2 + 0)$

Step 3.7

Add $2$ and $0$ .

$- {(2 x - 3)}^{- 1.5} + 0.5 {(2 x - 3)}^{- 0.5} \cdot 2$

Step 3.8

Multiply $2$ by $0.5$ .

$- {(2 x - 3)}^{- 1.5} + 1 {(2 x - 3)}^{- 0.5}$

Step 3.9

Multiply ${(2 x - 3)}^{- 0.5}$ by $1$ .

$- {(2 x - 3)}^{- 1.5} + {(2 x - 3)}^{- 0.5}$

Step 4

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{{(2 x - 3)}^{1.5}} + {(2 x - 3)}^{- 0.5}$

Step 5

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{{(2 x - 3)}^{1.5}} + \frac{1}{{(2 x - 3)}^{0.5}}$

Step 6

Reorder terms.

$\frac{1}{{(2 x - 3)}^{0.5}} - \frac{1}{{(2 x - 3)}^{1.5}}$