Calculus Examples

$\frac{1}{x + k e^{x}}$

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 $f (x) = 1$ and $g (x) = x + k e^{x}$ .

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

Step 2

Differentiate.

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Step 2.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.2

By the Sum Rule, the derivative of $x + k e^{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [k e^{x}]$ .

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

Step 2.3

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

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

Step 3

Evaluate $\frac{d}{d x} [k e^{x}]$ .

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Step 3.1

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

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

Step 3.2

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

$\frac{(x + k e^{x}) \cdot 0 - 1 \cdot 1 (1 + k e^{x})}{{(x + k e^{x})}^{2}}$

Step 4

Simplify.

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Step 4.1

Reorder terms.

$\frac{(x + k e^{x}) \cdot 0 - 1 \cdot 1 (e^{x} k + 1)}{{(x + k e^{x})}^{2}}$

Step 4.2

Reorder factors in $e^{x} k + 1$ .

$\frac{(x + k e^{x}) \cdot 0 - 1 \cdot 1 (k e^{x} + 1)}{{(x + k e^{x})}^{2}}$

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

$\frac{x \cdot 0 + k e^{x} \cdot 0 - 1 \cdot 1 (k e^{x} + 1)}{{(x + k e^{x})}^{2}}$

Step 5.2

Apply the distributive property.

$\frac{x \cdot 0 + k e^{x} \cdot 0 - 1 \cdot 1 (k e^{x}) - 1 \cdot 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3

Simplify the numerator.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

Multiply $x$ by $0$ .

$\frac{0 + k e^{x} \cdot 0 - 1 \cdot 1 (k e^{x}) - 1 \cdot 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3.1.2

Multiply $k e^{x} \cdot 0$ .

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Step 5.3.1.2.1

Multiply $0$ by $k$ .

$\frac{0 + 0 e^{x} - 1 \cdot 1 (k e^{x}) - 1 \cdot 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3.1.2.2

Multiply $0$ by $e^{x}$ .

$\frac{0 + 0 - 1 \cdot 1 (k e^{x}) - 1 \cdot 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3.1.3

Multiply $- 1$ by $1$ .

$\frac{0 + 0 - (k e^{x}) - 1 \cdot 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3.1.4

Multiply $- 1 \cdot 1 \cdot 1$ .

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Step 5.3.1.4.1

Multiply $- 1$ by $1$ .

$\frac{0 + 0 - k e^{x} - 1 \cdot 1}{{(x + k e^{x})}^{2}}$

Step 5.3.1.4.2

Multiply $- 1$ by $1$ .

$\frac{0 + 0 - k e^{x} - 1}{{(x + k e^{x})}^{2}}$

Step 5.3.2

Combine the opposite terms in $0 + 0 - k e^{x} - 1$ .

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Step 5.3.2.1

Add $0$ and $0$ .

$\frac{0 - k e^{x} - 1}{{(x + k e^{x})}^{2}}$

Step 5.3.2.2

Subtract $k e^{x}$ from $0$ .

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

Step 5.4

Reorder terms.

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

Step 5.5

Factor $- 1$ out of $- e^{x} k$ .

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

Step 5.6

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

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

Step 5.7

Factor $- 1$ out of $- (e^{x} k) - 1 (1)$ .

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

Step 5.8

Rewrite $- (e^{x} k + 1)$ as $- 1 (e^{x} k + 1)$ .

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

Step 5.9

Move the negative in front of the fraction.

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

Step 5.10

Reorder factors in $- \frac{e^{x} k + 1}{{(e^{x} k + x)}^{2}}$ .

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