Calculus Examples

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

Step 1

Move the negative in front of the fraction.

$\frac{d}{d x} [e^{- \frac{1}{x - 2}}]$

Step 2

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 $g (x) = - \frac{1}{x - 2}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $- \frac{1}{x - 2}$ .

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

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3

Replace all occurrences of $u_{1}$ with $- \frac{1}{x - 2}$ .

$e^{- \frac{1}{x - 2}} \frac{d}{d x} [- \frac{1}{x - 2}]$

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 $f (x) = - 1$ and $g (x) = \frac{1}{x - 2}$ .

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

Step 4

Rewrite $\frac{1}{x - 2}$ as ${(x - 2)}^{- 1}$ .

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

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x - 2$ .

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Step 5.1

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

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

Step 5.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

Multiply $- 1$ by $- 1$ .

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

Step 6.2

Multiply ${(x - 2)}^{- 2}$ by $1$ .

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

Step 6.3

By the Sum Rule, the derivative of $x - 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

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

Step 6.4

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

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

Step 6.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 6.6

Simplify the expression.

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Step 6.6.1

Add $1$ and $0$ .

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

Step 6.6.2

Multiply ${(x - 2)}^{- 2}$ by $1$ .

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

Step 6.7

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

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

Step 6.8

Simplify the expression.

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Step 6.8.1

Multiply $\frac{1}{x - 2}$ by $0$ .

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

Step 6.8.2

Add ${(x - 2)}^{- 2}$ and $0$ .

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

Step 7

Simplify.

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Step 7.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 7.2

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

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