Calculus Examples

$3 y^{3} = 5 x^{2} - 4 y - 3 y^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 y^{3}) = \frac{d}{d x} (5 x^{2} - 4 y - 3 y^{2})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 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 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Multiply $3$ by $3$ .

$9 (y^{2} \frac{d}{d x} [y])$

Step 2.4

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

$9 y^{2} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $5 x^{2} - 4 y - 3 y^{2}$ with respect to $x$ is $\frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- 3 y^{2}]$ .

$\frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- 3 y^{2}]$

Step 3.2

Evaluate $\frac{d}{d x} [5 x^{2}]$ .

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Step 3.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$5 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- 3 y^{2}]$

Step 3.2.2

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

$5 (2 x) + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- 3 y^{2}]$

Step 3.2.3

Multiply $2$ by $5$ .

$10 x + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- 3 y^{2}]$

Step 3.3

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

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Step 3.3.1

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

$10 x - 4 \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y^{2}]$

Step 3.3.2

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

$10 x - 4 y' + \frac{d}{d x} [- 3 y^{2}]$

Step 3.4

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

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Step 3.4.1

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

$10 x - 4 y' - 3 \frac{d}{d x} [y^{2}]$

Step 3.4.2

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 3.4.2.1

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

$10 x - 4 y' - 3 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

Step 3.4.2.2

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

$10 x - 4 y' - 3 (2 u \frac{d}{d x} [y])$

Step 3.4.2.3

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

$10 x - 4 y' - 3 (2 y \frac{d}{d x} [y])$

Step 3.4.3

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

$10 x - 4 y' - 3 (2 y y')$

Step 3.4.4

Multiply $2$ by $- 3$ .

$10 x - 4 y' - 6 y y'$

Step 3.5

Reorder terms.

$- 6 y y' + 10 x - 4 y'$

Step 4

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

$9 y^{2} y' = - 6 y y' + 10 x - 4 y'$

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.

$- 6 y y' + 10 x - 4 y' = 9 y^{2} y'$

Step 5.2

Subtract $9 y^{2} y'$ from both sides of the equation.

$- 6 y y' + 10 x - 4 y' - 9 y^{2} y' = 0$

Step 5.3

Subtract $10 x$ from both sides of the equation.

$- 6 y y' - 4 y' - 9 y^{2} y' = - 10 x$

Step 5.4

Factor $y'$ out of $- 6 y y' - 4 y' - 9 y^{2} y'$ .

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Step 5.4.1

Factor $y'$ out of $- 6 y y'$ .

$y' (- 6 y) - 4 y' - 9 y^{2} y' = - 10 x$

Step 5.4.2

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

$y' (- 6 y) + y' \cdot - 4 - 9 y^{2} y' = - 10 x$

Step 5.4.3

Factor $y'$ out of $- 9 y^{2} y'$ .

$y' (- 6 y) + y' \cdot - 4 + y' (- 9 y^{2}) = - 10 x$

Step 5.4.4

Factor $y'$ out of $y' (- 6 y) + y' \cdot - 4$ .

$y' (- 6 y - 4) + y' (- 9 y^{2}) = - 10 x$

Step 5.4.5

Factor $y'$ out of $y' (- 6 y - 4) + y' (- 9 y^{2})$ .

$y' (- 6 y - 4 - 9 y^{2}) = - 10 x$

Step 5.5

Reorder terms.

$y' (- 9 y^{2} - 6 y - 4) = - 10 x$

Step 5.6

Divide each term in $y' (- 9 y^{2} - 6 y - 4) = - 10 x$ by $- 9 y^{2} - 6 y - 4$ and simplify.

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Step 5.6.1

Divide each term in $y' (- 9 y^{2} - 6 y - 4) = - 10 x$ by $- 9 y^{2} - 6 y - 4$ .

$\frac{y' (- 9 y^{2} - 6 y - 4)}{- 9 y^{2} - 6 y - 4} = \frac{- 10 x}{- 9 y^{2} - 6 y - 4}$

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $- 9 y^{2} - 6 y - 4$ .

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Step 5.6.2.1.1

Cancel the common factor.

$\frac{y' (- 9 y^{2} - 6 y - 4)}{- 9 y^{2} - 6 y - 4} = \frac{- 10 x}{- 9 y^{2} - 6 y - 4}$

Step 5.6.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 10 x}{- 9 y^{2} - 6 y - 4}$

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Move the negative in front of the fraction.

$y' = - \frac{10 x}{- 9 y^{2} - 6 y - 4}$

Step 5.6.3.2

Factor $- 1$ out of $- 9 y^{2}$ .

$y' = - \frac{10 x}{- (9 y^{2}) - 6 y - 4}$

Step 5.6.3.3

Factor $- 1$ out of $- 6 y$ .

$y' = - \frac{10 x}{- (9 y^{2}) - (6 y) - 4}$

Step 5.6.3.4

Factor $- 1$ out of $- (9 y^{2}) - (6 y)$ .

$y' = - \frac{10 x}{- (9 y^{2} + 6 y) - 4}$

Step 5.6.3.5

Rewrite $- 4$ as $- 1 (4)$ .

$y' = - \frac{10 x}{- (9 y^{2} + 6 y) - 1 (4)}$

Step 5.6.3.6

Factor $- 1$ out of $- (9 y^{2} + 6 y) - 1 (4)$ .

$y' = - \frac{10 x}{- (9 y^{2} + 6 y + 4)}$

Step 5.6.3.7

Simplify the expression.

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Step 5.6.3.7.1

Rewrite $- (9 y^{2} + 6 y + 4)$ as $- 1 (9 y^{2} + 6 y + 4)$ .

$y' = - \frac{10 x}{- 1 (9 y^{2} + 6 y + 4)}$

Step 5.6.3.7.2

Move the negative in front of the fraction.

$y' = - - \frac{10 x}{9 y^{2} + 6 y + 4}$

Step 5.6.3.7.3

Multiply $- 1$ by $- 1$ .

$y' = 1 \frac{10 x}{9 y^{2} + 6 y + 4}$

Step 5.6.3.7.4

Multiply $\frac{10 x}{9 y^{2} + 6 y + 4}$ by $1$ .

$y' = \frac{10 x}{9 y^{2} + 6 y + 4}$

Step 6

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

$\frac{d y}{d x} = \frac{10 x}{9 y^{2} + 6 y + 4}$