Calculus Examples

$f (x, y) = \ln (x - y)$

Step 1

Find the first derivative.

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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 $g (x) = x - y$ .

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

$f' (x) = \frac{d}{d u} (\ln (u)) \frac{d}{d x} (x - y)$

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

$f' (x) = \frac{1}{u} \cdot \frac{d}{d x} (x - y)$

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

$f' (x) = \frac{1}{x - y} \cdot \frac{d}{d x} (x - y)$

Differentiate.

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

$f' (x) = \frac{1}{x - y} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (- y))$

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

$f' (x) = \frac{1}{x - y} \cdot (1 + \frac{d}{d x} (- y))$

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

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

Simplify the expression.

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Add $1$ and $0$ .

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

Multiply $\frac{1}{x - y}$ by $1$ .

$f' (x) = \frac{1}{x - y}$

Step 2

Find the second derivative.

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

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

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

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

$f'' (x) = \frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x - y)$

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

$f'' (x) = - u^{- 2} \frac{d}{d x} (x - y)$

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

$f'' (x) = - {(x - y)}^{- 2} \frac{d}{d x} (x - y)$

Differentiate.

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

$f'' (x) = - {(x - y)}^{- 2} (\frac{d}{d x} (x) + \frac{d}{d x} (- y))$

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

$f'' (x) = - {(x - y)}^{- 2} (1 + \frac{d}{d x} (- y))$

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

$f'' (x) = - {(x - y)}^{- 2} (1 + 0)$

Simplify the expression.

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Add $1$ and $0$ .

$f'' (x) = - {(x - y)}^{- 2} \cdot 1$

Multiply $- 1$ by $1$ .

$f'' (x) = - {(x - y)}^{- 2}$

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

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