Calculus Examples

\frac{d y}{d x} = 2 x^{3} y

$\frac{d y}{d x} = 2 x^{3} y$ ,

(1, 1)

$(1, 1)$

Step 1

Assume

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

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

Step 2

Check if the function is continuous in the neighborhood of

(1, 1)

$(1, 1)$ .

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Step 2.1

Substitute

(1, 1)

$(1, 1)$ values into

\frac{d y}{d x} = 2 x^{3} y

$\frac{d y}{d x} = 2 x^{3} y$ .

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Step 2.1.1

Substitute

1

$1$ for

x

$x$ .

2 \cdot 1^{3} y

$2 \cdot 1^{3} y$

Step 2.1.2

Substitute

1

$1$ for

y

$y$ .

2 \cdot 1^{3} \cdot 1

$2 \cdot 1^{3} \cdot 1$

2 \cdot 1^{3} \cdot 1

$2 \cdot 1^{3} \cdot 1$

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

$x$ value of

(1, 1)

$(1, 1)$ .

Continuous

Step 3

Find the partial derivative with respect to

y

$y$ .

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Step 3.1

Set up the partial derivative.

\frac{\partial f}{\partial y} = \frac{d}{d y} [2 x^{3} y]

$\frac{\partial f}{\partial y} = \frac{d}{d y} [2 x^{3} y]$

Step 3.2

Since

2 x^{3}

$2 x^{3}$ is constant with respect to

y

$y$ , the derivative of

2 x^{3} y

$2 x^{3} y$ with respect to

y

$y$ is

2 x^{3} \frac{d}{d y} [y]

$2 x^{3} \frac{d}{d y} [y]$ .

\frac{\partial f}{\partial y} = 2 x^{3} \frac{d}{d y} [y]

$\frac{\partial f}{\partial y} = 2 x^{3} \frac{d}{d y} [y]$

Step 3.3

Differentiate using the Power Rule which states that

\frac{d}{d y} [y^{n}]

$\frac{d}{d y} [y^{n}]$ is

n y^{n - 1}

$n y^{n - 1}$ where

n = 1

$n = 1$ .

\frac{\partial f}{\partial y} = 2 x^{3} \cdot 1

$\frac{\partial f}{\partial y} = 2 x^{3} \cdot 1$

Step 3.4

Multiply

2

$2$ by

1

$1$ .

\frac{\partial f}{\partial y} = 2 x^{3}

$\frac{\partial f}{\partial y} = 2 x^{3}$

\frac{\partial f}{\partial y} = 2 x^{3}

$\frac{\partial f}{\partial y} = 2 x^{3}$

Step 4

Check if the partial derivative with respect to

y

$y$ is continuous in the neighborhood of

(1, 1)

$(1, 1)$ .

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Step 4.1

y

$y$ value of

(1, 1)

$(1, 1)$ .

Continuous

Step 5

Both the function and its partial derivative with respect to

y

$y$ are continuous on an open interval around the

x

$x$ value of

(1, 1)

$(1, 1)$ .

One unique solution