Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $2 y^{3} - 2 - x y$ with respect to $x$ is $\frac{d}{d x} [2 y^{3}] + \frac{d}{d x} [- 2] + \frac{d}{d x} [- x y]$ .

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

Step 2.2

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

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Step 2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 y^{3}$ with respect to $x$ is $2 \frac{d}{d x} [y^{3}]$ .

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

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 $y$ .

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

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 $n = 3$ .

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $3$ by $2$ .

$6 y^{2} y' + \frac{d}{d x} [- 2] + \frac{d}{d x} [- x y]$

Step 2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$6 y^{2} y' + 0 + \frac{d}{d x} [- x y]$

Step 2.4

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

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Step 2.4.1

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

$6 y^{2} y' + 0 - \frac{d}{d x} [x y]$

Step 2.4.2

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 $g (x) = y$ .

$6 y^{2} y' + 0 - (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 2.4.3

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

$6 y^{2} y' + 0 - (x y' + y \frac{d}{d x} [x])$

Step 2.4.4

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

$6 y^{2} y' + 0 - (x y' + y \cdot 1)$

Step 2.4.5

Multiply $y$ by $1$ .

$6 y^{2} y' + 0 - (x y' + y)$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$6 y^{2} y' + 0 - (x y') - y$

Step 2.5.2

Add $6 y^{2} y'$ and $0$ .

$6 y^{2} y' - x y' - y$

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 $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

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

Step 3.2

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

$5 (3 x^{2})$

Step 3.3

Multiply $3$ by $5$ .

$15 x^{2}$

Step 4

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

$6 y^{2} y' - x y' - y = 15 x^{2}$

Step 5

Solve for $y'$ .

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Step 5.1

Add $y$ to both sides of the equation.

$6 y^{2} y' - x y' = 15 x^{2} + y$

Step 5.2

Factor $y'$ out of $6 y^{2} y' - x y'$ .

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Step 5.2.1

Factor $y'$ out of $6 y^{2} y'$ .

$y' (6 y^{2}) - x y' = 15 x^{2} + y$

Step 5.2.2

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

$y' (6 y^{2}) + y' (- x) = 15 x^{2} + y$

Step 5.2.3

Factor $y'$ out of $y' (6 y^{2}) + y' (- x)$ .

$y' (6 y^{2} - x) = 15 x^{2} + y$

Step 5.3

Divide each term in $y' (6 y^{2} - x) = 15 x^{2} + y$ by $6 y^{2} - x$ and simplify.

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Step 5.3.1

Divide each term in $y' (6 y^{2} - x) = 15 x^{2} + y$ by $6 y^{2} - x$ .

$\frac{y' (6 y^{2} - x)}{6 y^{2} - x} = \frac{15 x^{2}}{6 y^{2} - x} + \frac{y}{6 y^{2} - x}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $6 y^{2} - x$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (6 y^{2} - x)}{6 y^{2} - x} = \frac{15 x^{2}}{6 y^{2} - x} + \frac{y}{6 y^{2} - x}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{15 x^{2}}{6 y^{2} - x} + \frac{y}{6 y^{2} - x}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{15 x^{2} + y}{6 y^{2} - x}$

Step 6

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

$\frac{d y}{d x} = \frac{15 x^{2} + y}{6 y^{2} - x}$