Find the Derivative - d/dx y = natural log of e^(-x)+xe^(-x)

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

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

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

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

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

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

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

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

Step 2

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

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

Step 3

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 $g (x) = - x$ .

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

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

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

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

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

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

Step 4

Differentiate.

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

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

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

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

Simplify the expression.

Multiply $- 1$ by $1$ .

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

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

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

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

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

Step 5

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

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

Step 6

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 $g (x) = - x$ .

To apply the Chain Rule, set $u_{3}$ as $- x$ .

$\frac{1}{e^{- x} + x e^{- x}} (- e^{- x} + x (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x])$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$\frac{1}{e^{- x} + x e^{- x}} (- e^{- x} + x (e^{u_{3}} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x])$

Replace all occurrences of $u_{3}$ with $- x$ .

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

Step 7

Differentiate.

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

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

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

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

Simplify the expression.

Multiply $- 1$ by $1$ .

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

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

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

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

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

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

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

Simplify terms.

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

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

Add $- e^{- x}$ and $e^{- x}$ .

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

Add $x (- e^{- x})$ and $0$ .

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

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

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

Combine $\frac{x}{e^{- x} + x e^{- x}}$ and $e^{- x}$ .

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

Step 8

Simplify.

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

Multiply by $1$ .

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

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

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

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

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

Cancel the common factor of $e^{- x}$ .

Cancel the common factor.

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

Rewrite the expression.

$- \frac{x}{1 + x}$

Find the Derivative - d/dx y = natural log of e^(-x)+xe^(-x)

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