Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 2.2.2.1

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

$0 - (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 2.2.2.2

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

$0 - (3 u^{2} \frac{d}{d x} [y])$

Step 2.2.2.3

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

$0 - (3 y^{2} \frac{d}{d x} [y])$

Step 2.2.3

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

$0 - (3 y^{2} y')$

Step 2.2.4

Multiply $3$ by $- 1$ .

$0 - 3 y^{2} y'$

Step 2.3

Subtract $3 y^{2} y'$ from $0$ .

$- 3 y^{2} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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 $g (x) = y$ .

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

Step 3.3

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

$5 (x y' + y \frac{d}{d x} [x])$

Step 3.4

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

$5 (x y' + y \cdot 1)$

Step 3.5

Multiply $y$ by $1$ .

$5 (x y' + y)$

Step 3.6

Apply the distributive property.

$5 x y' + 5 y$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $5 x y'$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

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

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

Step 5.3.3.2

Factor $- 1$ out of $- 5 x$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Rewrite negatives.

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Step 5.3.3.4.1

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

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

Step 5.3.3.4.2

Move the negative in front of the fraction.

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

Step 6

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

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