Calculus Examples

$\frac{d y}{d x} = \sqrt{x - y}$ , $(3, 1)$

Step 1

Assume $\frac{d y}{d x} = f (x, y)$ .

Step 2

Check if the function is continuous in the neighborhood of $(3, 1)$ .

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Step 2.1

Substitute $(3, 1)$ values into $\frac{d y}{d x} = \sqrt{x - y}$ .

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Step 2.1.1

Substitute $3$ for $x$ .

$\sqrt{3 - y}$

Step 2.1.2

Substitute $1$ for $y$ .

$\sqrt{3 - 1}$

Step 2.1.3

Subtract $1$ from $3$ .

$\sqrt{2}$

Step 2.2

Since there is no log with negative or zero argument, no even radical with zero or negative radicand, and no fraction with zero in the denominator, the function is continuous on an open interval around the $x$ value of $(3, 1)$ .

Continuous

Step 3

Find the partial derivative with respect to $y$ .

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Step 3.1

Set up the partial derivative.

$\frac{\partial f}{\partial y} = \frac{d}{d y} [\sqrt{x - y}]$

Step 3.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - y}$ as ${(x - y)}^{\frac{1}{2}}$ .

$\frac{\partial f}{\partial y} = \frac{d}{d y} [{(x - y)}^{\frac{1}{2}}]$

Step 3.3

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

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Step 3.3.1

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

$\frac{\partial f}{\partial y} = \frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d y} [x - y]$

Step 3.3.2

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

$\frac{\partial f}{\partial y} = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d y} [x - y]$

Step 3.3.3

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

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{1}{2} - 1} \frac{d}{d y} [x - y]$

Step 3.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d y} [x - y]$

Step 3.5

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d y} [x - y]$

Step 3.6

Combine the numerators over the common denominator.

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d y} [x - y]$

Step 3.7

Simplify the numerator.

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Step 3.7.1

Multiply $- 1$ by $2$ .

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{1 - 2}{2}} \frac{d}{d y} [x - y]$

Step 3.7.2

Subtract $2$ from $1$ .

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{\frac{- 1}{2}} \frac{d}{d y} [x - y]$

Step 3.8

Combine fractions.

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Step 3.8.1

Move the negative in front of the fraction.

$\frac{\partial f}{\partial y} = \frac{1}{2} {(x - y)}^{- \frac{1}{2}} \frac{d}{d y} [x - y]$

Step 3.8.2

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

$\frac{\partial f}{\partial y} = \frac{{(x - y)}^{- \frac{1}{2}}}{2} \frac{d}{d y} [x - y]$

Step 3.8.3

Move ${(x - y)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\partial f}{\partial y} = \frac{1}{2 {(x - y)}^{\frac{1}{2}}} \frac{d}{d y} [x - y]$

Step 3.9

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

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

Step 3.10

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

$\frac{\partial f}{\partial y} = \frac{1}{2 {(x - y)}^{\frac{1}{2}}} (0 + \frac{d}{d y} [- y])$

Step 3.11

Add $0$ and $\frac{d}{d y} [- y]$ .

$\frac{\partial f}{\partial y} = \frac{1}{2 {(x - y)}^{\frac{1}{2}}} \frac{d}{d y} [- y]$

Step 3.12

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

$\frac{\partial f}{\partial y} = \frac{1}{2 {(x - y)}^{\frac{1}{2}}} (- \frac{d}{d y} [y])$

Step 3.13

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

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

Step 3.14

Combine fractions.

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Step 3.14.1

Multiply $- 1$ by $1$ .

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

Step 3.14.2

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

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

Step 3.14.3

Move the negative in front of the fraction.

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

Step 4

Check if the partial derivative with respect to $y$ is continuous in the neighborhood of $(3, 1)$ .

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Step 4.1

Convert fractional exponents to radicals.

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Step 4.1.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

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

Step 4.1.2

Anything raised to $1$ is the base itself.

$\frac{\partial f}{\partial y} = - \frac{1}{2 \sqrt{x - y}}$

Step 4.2

Substitute $(3, 1)$ values into $\frac{\partial f}{\partial y} = - \frac{1}{2 \sqrt{x - y}}$ .

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Step 4.2.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

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

Step 4.2.2

Anything raised to $1$ is the base itself.

$- \frac{1}{2 \sqrt{x - y}}$

Step 4.2.3

Substitute $3$ for $y$ .

$- \frac{1}{2 \sqrt{x - 3}}$

Step 4.3

Continuous

Step 5

Both the function and its partial derivative with respect to $y$ are continuous on an open interval around the $x$ value of $(3, 1)$ .

One unique solution