Calculus Examples

$x^{2} - 2 y^{3} + 4 y = 2$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} - 2 y^{3} + 4 y) = \frac{d}{d x} (2)$

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 $x^{2} - 2 y^{3} + 4 y$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 y^{3}] + \frac{d}{d x} [4 y]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 y^{3}] + \frac{d}{d x} [4 y]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [- 2 y^{3}] + \frac{d}{d x} [4 y]$

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

$2 x - 2 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [4 y]$

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 $y$ .

$2 x - 2 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [4 y]$

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 $n = 3$ .

$2 x - 2 (3 u^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [4 y]$

Step 2.2.2.3

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

$2 x - 2 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [4 y]$

Step 2.2.3

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

$2 x - 2 (3 y^{2} y') + \frac{d}{d x} [4 y]$

Step 2.2.4

Multiply $3$ by $- 2$ .

$2 x - 6 y^{2} y' + \frac{d}{d x} [4 y]$

Step 2.3

Evaluate $\frac{d}{d x} [4 y]$ .

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Step 2.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $4 \frac{d}{d x} [y]$ .

$2 x - 6 y^{2} y' + 4 \frac{d}{d x} [y]$

Step 2.3.2

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

$2 x - 6 y^{2} y' + 4 y'$

Step 2.4

Reorder terms.

$- 6 y^{2} y' + 2 x + 4 y'$

Step 3

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

$0$

Step 4

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

$- 6 y^{2} y' + 2 x + 4 y' = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $2 x$ from both sides of the equation.

$- 6 y^{2} y' + 4 y' = - 2 x$

Step 5.2

Factor $2 y'$ out of $- 6 y^{2} y' + 4 y'$ .

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Step 5.2.1

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

$2 y' (- 3 y^{2}) + 4 y' = - 2 x$

Step 5.2.2

Factor $2 y'$ out of $4 y'$ .

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

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

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Cancel the common factor of $- 2$ and $2$ .

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Step 5.3.3.1.1

Factor $2$ out of $- 2 x$ .

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

Step 5.3.3.1.2

Cancel the common factors.

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Step 5.3.3.1.2.1

Cancel the common factor.

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

Step 5.3.3.1.2.2

Rewrite the expression.

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

Step 5.3.3.2

Move the negative in front of the fraction.

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Rewrite $2$ as $- 1 (- 2)$ .

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

Step 5.3.3.5

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

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

Step 5.3.3.6

Simplify the expression.

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Step 5.3.3.6.1

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

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

Step 5.3.3.6.2

Move the negative in front of the fraction.

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

Step 5.3.3.6.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.6.4

Multiply $\frac{x}{3 y^{2} - 2}$ by $1$ .

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

Step 6

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

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