Calculus Examples

$- x y^{3} + 5 y = 3 - x^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- x y^{3} + 5 y) = \frac{d}{d x} (3 - x^{3})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $- x y^{3} + 5 y$ with respect to $x$ is $\frac{d}{d x} [- x y^{3}] + \frac{d}{d x} [5 y]$ .

$\frac{d}{d x} [- x y^{3}] + \frac{d}{d x} [5 y]$

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x y^{3}$ with respect to $x$ is $- \frac{d}{d x} [x y^{3}]$ .

$- \frac{d}{d x} [x y^{3}] + \frac{d}{d x} [5 y]$

Step 2.2.2

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

$- (x \frac{d}{d x} [y^{3}] + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 y]$

Step 2.2.3

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 2.2.3.1

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

$- (x (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 y]$

Step 2.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$- (x (3 u^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 y]$

Step 2.2.3.3

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

$- (x (3 y^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 y]$

Step 2.2.4

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$- (x (3 y^{2} y') + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [5 y]$

Step 2.2.5

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

$- (x (3 y^{2} y') + y^{3} \cdot 1) + \frac{d}{d x} [5 y]$

Step 2.2.6

Move $3$ to the left of $x$ .

$- (3 \cdot x y^{2} y' + y^{3} \cdot 1) + \frac{d}{d x} [5 y]$

Step 2.2.7

Multiply $y^{3}$ by $1$ .

$- (3 x y^{2} y' + y^{3}) + \frac{d}{d x} [5 y]$

Step 2.3

Evaluate $\frac{d}{d x} [5 y]$ .

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Step 2.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 y$ with respect to $x$ is $5 \frac{d}{d x} [y]$ .

$- (3 x y^{2} y' + y^{3}) + 5 \frac{d}{d x} [y]$

Step 2.3.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$- (3 x y^{2} y' + y^{3}) + 5 y'$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$- (3 x y^{2} y') - y^{3} + 5 y'$

Step 2.4.2

Multiply $3$ by $- 1$ .

$- 3 x y^{2} y' - y^{3} + 5 y'$

Step 2.4.3

Reorder terms.

$- y^{3} - 3 y^{2} x y' + 5 y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $3 - x^{3}$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [- x^{3}]$ .

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

Step 3.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- x^{3}]$

Step 3.2

Evaluate $\frac{d}{d x} [- x^{3}]$ .

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Step 3.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$0 - \frac{d}{d x} [x^{3}]$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$0 - (3 x^{2})$

Step 3.2.3

Multiply $3$ by $- 1$ .

$0 - 3 x^{2}$

Step 3.3

Subtract $3 x^{2}$ from $0$ .

$- 3 x^{2}$

Step 4

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

$- y^{3} - 3 y^{2} x y' + 5 y' = - 3 x^{2}$

Step 5

Solve for $y'$ .

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Step 5.1

Add $y^{3}$ to both sides of the equation.

$- 3 y^{2} x y' + 5 y' = - 3 x^{2} + y^{3}$

Step 5.2

Factor $y'$ out of $- 3 y^{2} x y' + 5 y'$ .

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Step 5.2.1

Factor $y'$ out of $- 3 y^{2} x y'$ .

$y' (- 3 y^{2} x) + 5 y' = - 3 x^{2} + y^{3}$

Step 5.2.2

Factor $y'$ out of $5 y'$ .

$y' (- 3 y^{2} x) + y' \cdot 5 = - 3 x^{2} + y^{3}$

Step 5.2.3

Factor $y'$ out of $y' (- 3 y^{2} x) + y' \cdot 5$ .

$y' (- 3 y^{2} x + 5) = - 3 x^{2} + y^{3}$

Step 5.3

Divide each term in $y' (- 3 y^{2} x + 5) = - 3 x^{2} + y^{3}$ by $- 3 y^{2} x + 5$ and simplify.

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Step 5.3.1

Divide each term in $y' (- 3 y^{2} x + 5) = - 3 x^{2} + y^{3}$ by $- 3 y^{2} x + 5$ .

$\frac{y' (- 3 y^{2} x + 5)}{- 3 y^{2} x + 5} = \frac{- 3 x^{2}}{- 3 y^{2} x + 5} + \frac{y^{3}}{- 3 y^{2} x + 5}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $- 3 y^{2} x + 5$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (- 3 y^{2} x + 5)}{- 3 y^{2} x + 5} = \frac{- 3 x^{2}}{- 3 y^{2} x + 5} + \frac{y^{3}}{- 3 y^{2} x + 5}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 3 x^{2}}{- 3 y^{2} x + 5} + \frac{y^{3}}{- 3 y^{2} x + 5}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 3 x^{2} + y^{3}}{- 3 y^{2} x + 5}$

Step 5.3.3.2

Factor $- 1$ out of $- 3 x^{2}$ .

$y' = \frac{- (3 x^{2}) + y^{3}}{- 3 y^{2} x + 5}$

Step 5.3.3.3

Factor $- 1$ out of $y^{3}$ .

$y' = \frac{- (3 x^{2}) - 1 (- y^{3})}{- 3 y^{2} x + 5}$

Step 5.3.3.4

Factor $- 1$ out of $- (3 x^{2}) - 1 (- y^{3})$ .

$y' = \frac{- (3 x^{2} - y^{3})}{- 3 y^{2} x + 5}$

Step 5.3.3.5

Rewrite $- (3 x^{2} - y^{3})$ as $- 1 (3 x^{2} - y^{3})$ .

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- 3 y^{2} x + 5}$

Step 5.3.3.6

Factor $- 1$ out of $- 3 y^{2} x$ .

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- (3 y^{2} x) + 5}$

Step 5.3.3.7

Rewrite $5$ as $- 1 (- 5)$ .

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- (3 y^{2} x) - 1 (- 5)}$

Step 5.3.3.8

Factor $- 1$ out of $- (3 y^{2} x) - 1 (- 5)$ .

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- (3 y^{2} x - 5)}$

Step 5.3.3.9

Rewrite $- (3 y^{2} x - 5)$ as $- 1 (3 y^{2} x - 5)$ .

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- 1 (3 y^{2} x - 5)}$

Step 5.3.3.10

Cancel the common factor.

$y' = \frac{- 1 (3 x^{2} - y^{3})}{- 1 (3 y^{2} x - 5)}$

Step 5.3.3.11

Rewrite the expression.

$y' = \frac{3 x^{2} - y^{3}}{3 y^{2} x - 5}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{3 x^{2} - y^{3}}{3 y^{2} x - 5}$