Calculus Examples

$y = \frac{\ln (x)}{e^{2 x}}$

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Multiply $2$ by $2$ .

$\frac{e^{2 x} \frac{d}{d x} [\ln (x)] - \ln (x) \frac{d}{d x} [e^{2 x}]}{e^{4 x}}$

Step 3

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

$\frac{e^{2 x} \frac{1}{x} - \ln (x) \frac{d}{d x} [e^{2 x}]}{e^{4 x}}$

Step 4

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

$\frac{\frac{e^{2 x}}{x} - \ln (x) \frac{d}{d x} [e^{2 x}]}{e^{4 x}}$

Step 5

Multiply $\frac{\frac{e^{2 x}}{x} - \ln (x) \frac{d}{d x} [e^{2 x}]}{e^{4 x}}$ by $\frac{x}{x}$ .

$\frac{x}{x} \cdot \frac{\frac{e^{2 x}}{x} - \ln (x) \frac{d}{d x} [e^{2 x}]}{e^{4 x}}$

Step 6

Simplify terms.

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Step 6.1

Combine.

$\frac{x (\frac{e^{2 x}}{x} - \ln (x) \frac{d}{d x} [e^{2 x}])}{x e^{4 x}}$

Step 6.2

Apply the distributive property.

$\frac{x \frac{e^{2 x}}{x} + x (- \ln (x) \frac{d}{d x} [e^{2 x}])}{x e^{4 x}}$

Step 6.3

Cancel the common factor of $x$ .

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Step 6.3.1

Cancel the common factor.

$\frac{x \frac{e^{2 x}}{x} + x (- \ln (x) \frac{d}{d x} [e^{2 x}])}{x e^{4 x}}$

Step 6.3.2

Rewrite the expression.

$\frac{e^{2 x} + x (- \ln (x) \frac{d}{d x} [e^{2 x}])}{x e^{4 x}}$

Step 7

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 $g (x) = 2 x$ .

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Step 7.1

To apply the Chain Rule, set $u$ as $2 x$ .

$\frac{e^{2 x} + x (- \ln (x) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x]))}{x e^{4 x}}$

Step 7.2

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

$\frac{e^{2 x} + x (- \ln (x) (e^{u} \frac{d}{d x} [2 x]))}{x e^{4 x}}$

Step 7.3

Replace all occurrences of $u$ with $2 x$ .

$\frac{e^{2 x} + x (- \ln (x) (e^{2 x} \frac{d}{d x} [2 x]))}{x e^{4 x}}$

Step 8

Differentiate.

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Step 8.1

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

$\frac{e^{2 x} + x (- \ln (x) (e^{2 x} (2 \frac{d}{d x} [x])))}{x e^{4 x}}$

Step 8.2

Multiply $2$ by $- 1$ .

$\frac{e^{2 x} + x (- 2 \ln (x) (e^{2 x} (\frac{d}{d x} [x])))}{x e^{4 x}}$

Step 8.3

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

$\frac{e^{2 x} + x (- 2 \ln (x) (e^{2 x} \cdot 1))}{x e^{4 x}}$

Step 8.4

Multiply $e^{2 x}$ by $1$ .

$\frac{e^{2 x} + x (- 2 \ln (x) e^{2 x})}{x e^{4 x}}$

Step 9

Simplify.

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Step 9.1

Simplify each term.

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Step 9.1.1

Rewrite using the commutative property of multiplication.

$\frac{e^{2 x} - 2 x (\ln (x) e^{2 x})}{x e^{4 x}}$

Step 9.1.2

Simplify $- 2 \ln (x)$ by moving $2$ inside the logarithm.

$\frac{e^{2 x} + x (- \ln (x^{2}) e^{2 x})}{x e^{4 x}}$

Step 9.1.3

Rewrite using the commutative property of multiplication.

$\frac{e^{2 x} - x (\ln (x^{2}) e^{2 x})}{x e^{4 x}}$

$\frac{e^{2 x} - x e^{2 x} \ln (x^{2})}{x e^{4 x}}$

Step 9.2

Reorder terms.

$\frac{- e^{2 x} x \ln (x^{2}) + e^{2 x}}{e^{4 x} x}$

Step 9.3

Factor $e^{2 x}$ out of $- e^{2 x} x \ln (x^{2}) + e^{2 x}$ .

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Step 9.3.1

Factor $e^{2 x}$ out of $- e^{2 x} x \ln (x^{2})$ .

$\frac{e^{2 x} (- x \ln (x^{2})) + e^{2 x}}{e^{4 x} x}$

Step 9.3.2

Multiply by $1$ .

$\frac{e^{2 x} (- x \ln (x^{2})) + e^{2 x} \cdot 1}{e^{4 x} x}$

Step 9.3.3

Factor $e^{2 x}$ out of $e^{2 x} (- x \ln (x^{2})) + e^{2 x} \cdot 1$ .

$\frac{e^{2 x} (- x \ln (x^{2}) + 1)}{e^{4 x} x}$

Step 9.4

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

$\frac{e^{2 x} (- x (2 \ln (x)) + 1)}{e^{4 x} x}$

Step 9.5

Cancel the common factor of $e^{2 x}$ and $e^{4 x}$ .

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Step 9.5.1

Factor $e^{4 x}$ out of $e^{2 x} (- x (2 \ln (x)) + 1)$ .

$\frac{e^{4 x} (e^{- 2 x} (- x (2 \ln (x)) + 1))}{e^{4 x} x}$

Step 9.5.2

Cancel the common factors.

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Step 9.5.2.1

Factor $e^{4 x}$ out of $e^{4 x} x$ .

$\frac{e^{4 x} (e^{- 2 x} (- x (2 \ln (x)) + 1))}{e^{4 x} (x)}$

Step 9.5.2.2

Cancel the common factor.

$\frac{e^{4 x} (e^{- 2 x} (- x (2 \ln (x)) + 1))}{e^{4 x} x}$

Step 9.5.2.3

Rewrite the expression.

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

Step 9.6

Multiply $2$ by $- 1$ .

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

Step 9.7

Factor $- 1$ out of $- 2 x \ln (x)$ .

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

Step 9.8

Rewrite $1$ as $- 1 (- 1)$ .

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

Step 9.9

Factor $- 1$ out of $- (2 x \ln (x)) - 1 (- 1)$ .

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

Step 9.10

Rewrite $- (2 x \ln (x) - 1)$ as $- 1 (2 x \ln (x) - 1)$ .

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

Step 9.11

Move the negative in front of the fraction.

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