Calculus Examples

$f (x) = 7 x^{4} x^{\frac{1}{2}} + (- \frac{8}{x^{2} x^{0.5}})$

Step 1

By the Sum Rule, the derivative of $7 x^{4} x^{\frac{1}{2}} + (- \frac{8}{x^{2} x^{0.5}})$ with respect to $x$ is $\frac{d}{d x} [7 x^{4} x^{\frac{1}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$ .

$\frac{d}{d x} [7 x^{4} x^{\frac{1}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2

Evaluate $\frac{d}{d x} [7 x^{4} x^{\frac{1}{2}}]$ .

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Step 2.1

Multiply $x^{4}$ by $x^{\frac{1}{2}}$ by adding the exponents.

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Step 2.1.1

Move $x^{\frac{1}{2}}$ .

$\frac{d}{d x} [7 (x^{\frac{1}{2}} x^{4})] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.2

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

$\frac{d}{d x} [7 x^{\frac{1}{2} + 4}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.3

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{d}{d x} [7 x^{\frac{1}{2} + 4 \cdot \frac{2}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.4

Combine $4$ and $\frac{2}{2}$ .

$\frac{d}{d x} [7 x^{\frac{1}{2} + \frac{4 \cdot 2}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.5

Combine the numerators over the common denominator.

$\frac{d}{d x} [7 x^{\frac{1 + 4 \cdot 2}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.6

Simplify the numerator.

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Step 2.1.6.1

Multiply $4$ by $2$ .

$\frac{d}{d x} [7 x^{\frac{1 + 8}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.1.6.2

Add $1$ and $8$ .

$\frac{d}{d x} [7 x^{\frac{9}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.2

Since $7$ is constant with respect to $x$ , the derivative of $7 x^{\frac{9}{2}}$ with respect to $x$ is $7 \frac{d}{d x} [x^{\frac{9}{2}}]$ .

$7 \frac{d}{d x} [x^{\frac{9}{2}}] + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = \frac{9}{2}$ .

$7 (\frac{9}{2} x^{\frac{9}{2} - 1}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.4

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

$7 (\frac{9}{2} x^{\frac{9}{2} - 1 \cdot \frac{2}{2}}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.5

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

$7 (\frac{9}{2} x^{\frac{9}{2} + \frac{- 1 \cdot 2}{2}}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.6

Combine the numerators over the common denominator.

$7 (\frac{9}{2} x^{\frac{9 - 1 \cdot 2}{2}}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.7

Simplify the numerator.

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Step 2.7.1

Multiply $- 1$ by $2$ .

$7 (\frac{9}{2} x^{\frac{9 - 2}{2}}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.7.2

Subtract $2$ from $9$ .

$7 (\frac{9}{2} x^{\frac{7}{2}}) + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.8

Combine $\frac{9}{2}$ and $x^{\frac{7}{2}}$ .

$7 \frac{9 x^{\frac{7}{2}}}{2} + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.9

Combine $7$ and $\frac{9 x^{\frac{7}{2}}}{2}$ .

$\frac{7 (9 x^{\frac{7}{2}})}{2} + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 2.10

Multiply $9$ by $7$ .

$\frac{63 x^{\frac{7}{2}}}{2} + \frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$

Step 3

Evaluate $\frac{d}{d x} [- \frac{8}{x^{2} x^{0.5}}]$ .

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Step 3.1

Multiply $x^{2}$ by $x^{0.5}$ by adding the exponents.

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Step 3.1.1

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

$\frac{63 x^{\frac{7}{2}}}{2} + \frac{d}{d x} [- \frac{8}{x^{2 + 0.5}}]$

Step 3.1.2

Add $2$ and $0.5$ .

$\frac{63 x^{\frac{7}{2}}}{2} + \frac{d}{d x} [- \frac{8}{x^{2.5}}]$

Step 3.2

Since $- 8$ is constant with respect to $x$ , the derivative of $- \frac{8}{x^{2.5}}$ with respect to $x$ is $- 8 \frac{d}{d x} [\frac{1}{x^{2.5}}]$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 \frac{d}{d x} [\frac{1}{x^{2.5}}]$

Step 3.3

Rewrite $\frac{1}{x^{2.5}}$ as ${(x^{2.5})}^{- 1}$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 \frac{d}{d x} [{(x^{2.5})}^{- 1}]$

Step 3.4

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 $g (x) = x^{2.5}$ .

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Step 3.4.1

To apply the Chain Rule, set $u$ as $x^{2.5}$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2.5}])$

Step 3.4.2

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

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- u^{- 2} \frac{d}{d x} [x^{2.5}])$

Step 3.4.3

Replace all occurrences of $u$ with $x^{2.5}$ .

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Step 3.4.3.1

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

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- {(x^{2 + 0.5})}^{- 2} \frac{d}{d x} [x^{2.5}])$

Step 3.4.3.2

Add $2$ and $0.5$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- {(x^{2.5})}^{- 2} \frac{d}{d x} [x^{2.5}])$

Step 3.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2.5$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- {(x^{2.5})}^{- 2} (2.5 x^{1.5}))$

Step 3.6

Multiply the exponents in ${(x^{2.5})}^{- 2}$ .

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Step 3.6.1

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

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- x^{2.5 \cdot - 2} (2.5 x^{1.5}))$

Step 3.6.2

Multiply $2.5$ by $- 2$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- x^{- 5} (2.5 x^{1.5}))$

Step 3.7

Multiply $2.5$ by $- 1$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- 2.5 x^{- 5} x^{1.5})$

Step 3.8

Multiply $x^{- 5}$ by $x^{1.5}$ by adding the exponents.

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Step 3.8.1

Move $x^{1.5}$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- 2.5 (x^{1.5} x^{- 5}))$

Step 3.8.2

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

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- 2.5 x^{1.5 - 5})$

Step 3.8.3

Subtract $5$ from $1.5$ .

$\frac{63 x^{\frac{7}{2}}}{2} - 8 (- 2.5 x^{- 3.5})$

Step 3.9

Multiply $- 2.5$ by $- 8$ .

$\frac{63 x^{\frac{7}{2}}}{2} + 20 x^{- 3.5}$

Step 4

Simplify.

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Step 4.1

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

$\frac{63 x^{\frac{7}{2}}}{2} + 20 \frac{1}{x^{3.5}}$

Step 4.2

Combine $20$ and $\frac{1}{x^{3.5}}$ .

$\frac{63 x^{\frac{7}{2}}}{2} + \frac{20}{x^{3.5}}$