Calculus Examples

$y = \ln (1 - x e^{- x})$

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 $g (x) = 1 - x e^{- x}$ .

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Step 1.1

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

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

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [1 - x e^{- x}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $1 - x e^{- x}$ .

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 3

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 $f (x) = x$ and $g (x) = e^{- x}$ .

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

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 $g (x) = - x$ .

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Step 4.1

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

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

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 $a$ = $e$ .

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

Step 4.3

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

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the expression.

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Step 5.3.1

Multiply $- 1$ by $1$ .

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

Step 5.3.2

Move $- 1$ to the left of $e^{- x}$ .

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

Step 5.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 5.4

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

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

Step 5.5

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

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

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

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

Step 6.2

Combine terms.

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Step 6.2.1

Multiply $- 1$ by $- 1$ .

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

Step 6.2.2

Multiply $x$ by $1$ .

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

Step 6.3

Reorder the factors of $\frac{1}{1 - x e^{- x}} (x e^{- x} - e^{- x})$ .

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

Step 6.4

Multiply $x e^{- x} - e^{- x}$ by $\frac{1}{1 - x e^{- x}}$ .

$\frac{x e^{- x} - e^{- x}}{1 - x e^{- x}}$

Step 6.5

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

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Step 6.5.1

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

$\frac{e^{- x} x - e^{- x}}{1 - x e^{- x}}$

Step 6.5.2

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

$\frac{e^{- x} x + e^{- x} \cdot - 1}{1 - x e^{- x}}$

Step 6.5.3

Factor $e^{- x}$ out of $e^{- x} x + e^{- x} \cdot - 1$ .

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