Calculus Examples

$y = x e^{- x}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} (x e^{- x})$

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

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

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

Step 3.4

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

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

Step 3.5

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$x (e^{- x} (- x')) + e^{- x} x'$

Step 3.6

Simplify.

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Step 3.6.1

Reorder terms.

$- e^{- x} x x' + e^{- x} x'$

Step 3.6.2

Reorder factors in $- e^{- x} x x' + e^{- x} x'$ .

$- x e^{- x} x' + e^{- x} x'$

Step 4

Reform the equation by setting the left side equal to the right side.

$1 = - x e^{- x} x' + e^{- x} x'$

Step 5

Solve for $x'$ .

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Step 5.1

Rewrite the equation as $- x e^{- x} x' + e^{- x} x' = 1$ .

$- x e^{- x} x' + e^{- x} x' = 1$

Step 5.2

Reorder factors in $- x e^{- x} x' + e^{- x} x'$ .

$- x x' e^{- x} + x' e^{- x} = 1$

Step 5.3

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

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Step 5.3.1

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

$x' e^{- x} (- x) + x' e^{- x} = 1$

Step 5.3.2

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

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

Step 5.3.3

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

$x' e^{- x} (- x + 1) = 1$

Step 5.4

Divide each term in $x' e^{- x} (- x + 1) = 1$ by $e^{- x} (- x + 1)$ and simplify.

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Step 5.4.1

Divide each term in $x' e^{- x} (- x + 1) = 1$ by $e^{- x} (- x + 1)$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

Cancel the common factor of $- x + 1$ .

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $x'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Factor $- 1$ out of $- x$ .

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

Step 5.4.3.2

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

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

Step 5.4.3.3

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

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

Step 5.4.3.4

Rewrite negatives.

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Step 5.4.3.4.1

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

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

Step 5.4.3.4.2

Move the negative in front of the fraction.

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

Step 6

Replace $x'$ with $\frac{d x}{d y}$ .

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