Calculus Examples

y = x^{2} y^{3} + x^{3} y^{2}

$y = x^{2} y^{3} + x^{3} y^{2}$

Step 1

Differentiate both sides of the equation.

\frac{d}{d x} (y) = \frac{d}{d x} (x^{2} y^{3} + x^{3} y^{2})

$\frac{d}{d x} (y) = \frac{d}{d x} (x^{2} y^{3} + x^{3} y^{2})$

Step 2

The derivative of

y

$y$ with respect to

x

$x$ is

$y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of

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

$x$ is

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

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

Step 3.2

Evaluate

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

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Step 3.2.1

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

$f (x) = x^{2}$ and

$g (x) = y^{3}$ .

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

Step 3.2.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = x^{3}$ and

$g (x) = y$ .

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Step 3.2.2.1

To apply the Chain Rule, set

$u_{1}$ as

$y$ .

$x^{2} (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.2.2

Differentiate using the Power Rule which states that

$\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is

$n {u_{1}}^{n - 1}$ where

$n = 3$ .

$x^{2} (3 {u_{1}}^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.2.3

Replace all occurrences of

$u_{1}$ with

$y$ .

$x^{2} (3 y^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.3

Rewrite

$\frac{d}{d x} [y]$ as

$y'$ .

$x^{2} (3 y^{2} y') + y^{3} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.4

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 2$ .

$x^{2} (3 y^{2} y') + y^{3} (2 x) + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.5

Move

$3$ to the left of

$x^{2}$ .

$3 \cdot x^{2} y^{2} y' + y^{3} (2 x) + \frac{d}{d x} [x^{3} y^{2}]$

Step 3.2.6

Move

$2$ to the left of

$y^{3}$ .

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

Step 3.3

Evaluate

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

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Step 3.3.1

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

$f (x) = x^{3}$ and

$g (x) = y^{2}$ .

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

Step 3.3.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = x^{2}$ and

$g (x) = y$ .

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Step 3.3.2.1

To apply the Chain Rule, set

$u_{2}$ as

$y$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + x^{3} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{3}]$

Step 3.3.2.2

Differentiate using the Power Rule which states that

$\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is

$n {u_{2}}^{n - 1}$ where

$n = 2$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + x^{3} (2 u_{2} \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{3}]$

Step 3.3.2.3

Replace all occurrences of

$u_{2}$ with

$y$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + x^{3} (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{3}]$

Step 3.3.3

Rewrite

$\frac{d}{d x} [y]$ as

$y'$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + x^{3} (2 y y') + y^{2} \frac{d}{d x} [x^{3}]$

Step 3.3.4

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 3$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + x^{3} (2 y y') + y^{2} (3 x^{2})$

Step 3.3.5

Move

$2$ to the left of

$x^{3}$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + 2 \cdot x^{3} y y' + y^{2} (3 x^{2})$

Step 3.3.6

Move

$3$ to the left of

$y^{2}$ .

$3 x^{2} y^{2} y' + 2 y^{3} x + 2 x^{3} y y' + 3 y^{2} x^{2}$

Step 3.4

Reorder terms.

$3 x^{2} y^{2} y' + 2 y^{3} x + 2 x^{3} y y' + 3 x^{2} y^{2}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = 3 x^{2} y^{2} y' + 2 y^{3} x + 2 x^{3} y y' + 3 x^{2} y^{2}$

Step 5

Solve for

$y'$ .

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Step 5.1

Since

$y'$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} y^{2} y' + 2 y^{3} x + 2 x^{3} y y' + 3 x^{2} y^{2} = y'$

Step 5.2

Subtract

$y'$ from both sides of the equation.

$3 x^{2} y^{2} y' + 2 y^{3} x + 2 x^{3} y y' + 3 x^{2} y^{2} - y' = 0$

Step 5.3

Move all terms not containing

$y'$ to the right side of the equation.

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Step 5.3.1

Subtract

$2 y^{3} x$ from both sides of the equation.

$3 x^{2} y^{2} y' + 2 x^{3} y y' + 3 x^{2} y^{2} - y' = - 2 y^{3} x$

Step 5.3.2

Subtract

$3 x^{2} y^{2}$ from both sides of the equation.

$3 x^{2} y^{2} y' + 2 x^{3} y y' - y' = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.4

Factor

$y'$ out of

$3 x^{2} y^{2} y' + 2 x^{3} y y' - y'$ .

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Step 5.4.1

Factor

$y'$ out of

$3 x^{2} y^{2} y'$ .

$y' (3 x^{2} y^{2}) + 2 x^{3} y y' - y' = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.4.2

Factor

$y'$ out of

$2 x^{3} y y'$ .

$y' (3 x^{2} y^{2}) + y' (2 x^{3} y) - y' = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.4.3

Factor

$y'$ out of

$- y'$ .

$y' (3 x^{2} y^{2}) + y' (2 x^{3} y) + y' \cdot - 1 = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.4.4

Factor

$y'$ out of

$y' (3 x^{2} y^{2}) + y' (2 x^{3} y)$ .

$y' (3 x^{2} y^{2} + 2 x^{3} y) + y' \cdot - 1 = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.4.5

Factor

$y'$ out of

$y' (3 x^{2} y^{2} + 2 x^{3} y) + y' \cdot - 1$ .

$y' (3 x^{2} y^{2} + 2 x^{3} y - 1) = - 2 y^{3} x - 3 x^{2} y^{2}$

Step 5.5

Divide each term in

$y' (3 x^{2} y^{2} + 2 x^{3} y - 1) = - 2 y^{3} x - 3 x^{2} y^{2}$ by

$3 x^{2} y^{2} + 2 x^{3} y - 1$ and simplify.

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Step 5.5.1

Divide each term in

$y' (3 x^{2} y^{2} + 2 x^{3} y - 1) = - 2 y^{3} x - 3 x^{2} y^{2}$ by

$3 x^{2} y^{2} + 2 x^{3} y - 1$ .

$\frac{y' (3 x^{2} y^{2} + 2 x^{3} y - 1)}{3 x^{2} y^{2} + 2 x^{3} y - 1} = \frac{- 2 y^{3} x}{3 x^{2} y^{2} + 2 x^{3} y - 1} + \frac{- 3 x^{2} y^{2}}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of

$3 x^{2} y^{2} + 2 x^{3} y - 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' (3 x^{2} y^{2} + 2 x^{3} y - 1)}{3 x^{2} y^{2} + 2 x^{3} y - 1} = \frac{- 2 y^{3} x}{3 x^{2} y^{2} + 2 x^{3} y - 1} + \frac{- 3 x^{2} y^{2}}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.2.1.2

Divide

$y'$ by

$1$ .

$y' = \frac{- 2 y^{3} x}{3 x^{2} y^{2} + 2 x^{3} y - 1} + \frac{- 3 x^{2} y^{2}}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 2 y^{3} x - 3 x^{2} y^{2}}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.2

Factor

$y^{2} x$ out of

$- 2 y^{3} x - 3 x^{2} y^{2}$ .

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Step 5.5.3.2.1

Factor

$y^{2} x$ out of

$- 2 y^{3} x$ .

$y' = \frac{y^{2} x (- 2 y) - 3 x^{2} y^{2}}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.2.2

Factor

$y^{2} x$ out of

$- 3 x^{2} y^{2}$ .

$y' = \frac{y^{2} x (- 2 y) + y^{2} x (- 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.2.3

Factor

$y^{2} x$ out of

$y^{2} x (- 2 y) + y^{2} x (- 3 x)$ .

$y' = \frac{y^{2} x (- 2 y - 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.3

Factor

$- 1$ out of

$- 2 y$ .

$y' = \frac{y^{2} x (- (2 y) - 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.4

Factor

$- 1$ out of

$- 3 x$ .

$y' = \frac{y^{2} x (- (2 y) - (3 x))}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.5

Factor

$- 1$ out of

$- (2 y) - (3 x)$ .

$y' = \frac{y^{2} x (- (2 y + 3 x))}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.6

Simplify the expression.

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Step 5.5.3.6.1

Rewrite

$- (2 y + 3 x)$ as

$- 1 (2 y + 3 x)$ .

$y' = \frac{y^{2} x (- 1 (2 y + 3 x))}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.6.2

Move the negative in front of the fraction.

$y' = - \frac{(y^{2} x) (2 y + 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 5.5.3.6.3

Reorder factors in

$- \frac{(y^{2} x) (2 y + 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$ .

$y' = - \frac{y^{2} x (2 y + 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

Step 6

Replace

$y'$ with

$\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{y^{2} x (2 y + 3 x)}{3 x^{2} y^{2} + 2 x^{3} y - 1}$

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