Calculus Examples

$f (x) = e^{- x} \ln (x)$ , $(1, 0)$

Step 1

Find the first derivative and evaluate at $x = 1$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

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 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

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

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

Step 1.5

Differentiate.

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Step 1.5.1

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

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

Step 1.5.2

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

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

Step 1.5.3

Simplify the expression.

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Step 1.5.3.1

Multiply $- 1$ by $1$ .

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

Step 1.5.3.2

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

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

Step 1.5.3.3

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

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

Step 1.5.3.4

Reorder terms.

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

Step 1.6

Evaluate the derivative at $x = 1$ .

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

Step 1.7

Simplify.

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Step 1.7.1

Multiply $- 1$ by $1$ .

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

Step 1.7.2

Simplify each term.

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Step 1.7.2.1

Multiply $- 1$ by $1$ .

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

Step 1.7.2.2

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

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

Step 1.7.2.3

The natural logarithm of $1$ is $0$ .

$- \frac{1}{e} \cdot 0 + \frac{e^{- 1}}{1}$

Step 1.7.2.4

Multiply $- \frac{1}{e} \cdot 0$ .

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Step 1.7.2.4.1

Multiply $0$ by $- 1$ .

$0 \frac{1}{e} + \frac{e^{- 1}}{1}$

Step 1.7.2.4.2

Multiply $0$ by $\frac{1}{e}$ .

$0 + \frac{e^{- 1}}{1}$

Step 1.7.2.5

Move $e^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$0 + \frac{1}{e}$

Step 1.7.3

Add $0$ and $\frac{1}{e}$ .

$\frac{1}{e}$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $\frac{1}{e}$ and a given point $(1, 0)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = \frac{1}{e} \cdot (x - (1))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y + 0 = \frac{1}{e} \cdot (x - 1)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

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

Step 2.3.2

Simplify $\frac{1}{e} \cdot (x - 1)$ .

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Step 2.3.2.1

Apply the distributive property.

$y = \frac{1}{e} x + \frac{1}{e} \cdot - 1$

Step 2.3.2.2

Combine $\frac{1}{e}$ and $x$ .

$y = \frac{x}{e} + \frac{1}{e} \cdot - 1$

Step 2.3.2.3

Combine $\frac{1}{e}$ and $- 1$ .

$y = \frac{x}{e} + \frac{- 1}{e}$

Step 2.3.2.4

Move the negative in front of the fraction.

$y = \frac{x}{e} - \frac{1}{e}$

Step 2.3.3

Reorder terms.

$y = \frac{1}{e} x - \frac{1}{e}$

Step 3