Calculus Examples

$y = \ln (10 x^{2 - 5 x})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 10 x^{2 - 5 x}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $10 x^{2 - 5 x}$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [10 x^{2 - 5 x}]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [10 x^{2 - 5 x}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $10 x^{2 - 5 x}$ .

$\frac{1}{10 x^{2 - 5 x}} \frac{d}{d x} [10 x^{2 - 5 x}]$

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x^{2 - 5 x}$ with respect to $x$ is $10 \frac{d}{d x} [x^{2 - 5 x}]$ .

$\frac{1}{10 x^{2 - 5 x}} (10 \frac{d}{d x} [x^{2 - 5 x}])$

Step 2.2

Simplify terms.

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Step 2.2.1

Combine $10$ and $\frac{1}{10 x^{2 - 5 x}}$ .

$\frac{10}{10 x^{2 - 5 x}} \frac{d}{d x} [x^{2 - 5 x}]$

Step 2.2.2

Cancel the common factor of $10$ .

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Step 2.2.2.1

Cancel the common factor.

$\frac{10}{10 x^{2 - 5 x}} \frac{d}{d x} [x^{2 - 5 x}]$

Step 2.2.2.2

Rewrite the expression.

$\frac{1}{x^{2 - 5 x}} \frac{d}{d x} [x^{2 - 5 x}]$

Step 3

Use the properties of logarithms to simplify the differentiation.

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Step 3.1

Rewrite $x^{2 - 5 x}$ as $e^{\ln (x^{2 - 5 x})}$ .

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

Step 3.2

Expand $\ln (x^{2 - 5 x})$ by moving $2 - 5 x$ outside the logarithm.

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

Step 4

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

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $(2 - 5 x) \ln (x)$ .

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

Step 4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 4.3

Replace all occurrences of $u_{2}$ with $(2 - 5 x) \ln (x)$ .

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

Step 5

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

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

Step 6

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 2 - 5 x$ and $g (x) = \ln (x)$ .

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

Step 7

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

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

Step 8

Differentiate.

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Step 8.1

By the Sum Rule, the derivative of $2 - 5 x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [- 5 x]$ .

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

Step 8.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{e^{(2 - 5 x) \ln (x)}}{x^{2 - 5 x}} ((2 - 5 x) \frac{1}{x} + \ln (x) (0 + \frac{d}{d x} [- 5 x]))$

Step 8.3

Add $0$ and $\frac{d}{d x} [- 5 x]$ .

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

Step 8.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

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

Step 8.5

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

$\frac{e^{(2 - 5 x) \ln (x)}}{x^{2 - 5 x}} ((2 - 5 x) \frac{1}{x} + \ln (x) (- 5 \cdot 1))$

Step 8.6

Simplify the expression.

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Step 8.6.1

Multiply $- 5$ by $1$ .

$\frac{e^{(2 - 5 x) \ln (x)}}{x^{2 - 5 x}} ((2 - 5 x) \frac{1}{x} + \ln (x) \cdot - 5)$

Step 8.6.2

Move $- 5$ to the left of $\ln (x)$ .

$\frac{e^{(2 - 5 x) \ln (x)}}{x^{2 - 5 x}} ((2 - 5 x) \frac{1}{x} - 5 \cdot \ln (x))$

Step 9

Simplify.

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Step 9.1

Apply the distributive property.

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} ((2 - 5 x) \frac{1}{x} - 5 \ln (x))$

Step 9.2

Apply the distributive property.

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (2 \frac{1}{x} - 5 x \frac{1}{x} - 5 \ln (x))$

Step 9.3

Combine terms.

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Step 9.3.1

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

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} - 5 x \frac{1}{x} - 5 \ln (x))$

Step 9.3.2

Combine $- 5$ and $\frac{1}{x}$ .

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} + x \frac{- 5}{x} - 5 \ln (x))$

Step 9.3.3

Combine $x$ and $\frac{- 5}{x}$ .

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} + \frac{x \cdot - 5}{x} - 5 \ln (x))$

Step 9.3.4

Move $- 5$ to the left of $x$ .

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} + \frac{- 5 \cdot x}{x} - 5 \ln (x))$

Step 9.3.5

Cancel the common factor of $x$ .

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Step 9.3.5.1

Cancel the common factor.

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} + \frac{- 5 x}{x} - 5 \ln (x))$

Step 9.3.5.2

Divide $- 5$ by $1$ .

$\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} - 5 - 5 \ln (x))$

Step 9.4

Reorder the factors of $\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}} (\frac{2}{x} - 5 - 5 \ln (x))$ .

$(\frac{2}{x} - 5 - 5 \ln (x)) \frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.5

Multiply $\frac{2}{x} - 5 - 5 \ln (x)$ by $\frac{e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$ .

$\frac{(\frac{2}{x} - 5 - 5 \ln (x)) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6

Simplify the numerator.

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Step 9.6.1

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{(\frac{2}{x} - 5 \cdot \frac{x}{x} - 5 \ln (x)) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6.2

Combine $- 5$ and $\frac{x}{x}$ .

$\frac{(\frac{2}{x} + \frac{- 5 x}{x} - 5 \ln (x)) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6.3

Combine the numerators over the common denominator.

$\frac{(\frac{2 - 5 x}{x} - 5 \ln (x)) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6.4

To write $- 5 \ln (x)$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{(\frac{2 - 5 x}{x} - 5 \ln (x) \cdot \frac{x}{x}) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6.5

Combine $- 5 \ln (x)$ and $\frac{x}{x}$ .

$\frac{(\frac{2 - 5 x}{x} + \frac{- 5 \ln (x) x}{x}) e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.6.6

Combine the numerators over the common denominator.

$\frac{\frac{2 - 5 x - 5 \ln (x) x}{x} e^{2 \ln (x) - 5 x \ln (x)}}{x^{2 - 5 x}}$

Step 9.7

Combine $\frac{2 - 5 x - 5 \ln (x) x}{x}$ and $e^{2 \ln (x) - 5 x \ln (x)}$ .

$\frac{\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)}}{x}}{x^{2 - 5 x}}$

Step 9.8

Multiply the numerator by the reciprocal of the denominator.

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)}}{x} \cdot \frac{1}{x^{2 - 5 x}}$

Step 9.9

Combine.

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)} \cdot 1}{x \cdot x^{2 - 5 x}}$

Step 9.10

Multiply $x$ by $x^{2 - 5 x}$ by adding the exponents.

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Step 9.10.1

Multiply $x$ by $x^{2 - 5 x}$ .

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Step 9.10.1.1

Raise $x$ to the power of $1$ .

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)} \cdot 1}{x^{1} x^{2 - 5 x}}$

Step 9.10.1.2

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

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)} \cdot 1}{x^{1 + 2 - 5 x}}$

Step 9.10.2

Add $1$ and $2$ .

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)} \cdot 1}{x^{3 - 5 x}}$

Step 9.11

Multiply $(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)}$ by $1$ .

$\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)}}{x^{3 - 5 x}}$

Step 9.12

Reorder factors in $\frac{(2 - 5 x - 5 \ln (x) x) e^{2 \ln (x) - 5 x \ln (x)}}{x^{3 - 5 x}}$ .

$\frac{e^{2 \ln (x) - 5 x \ln (x)} (2 - 5 x - 5 x \ln (x))}{x^{3 - 5 x}}$