Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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

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

Step 2.1.2

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

$0 + \frac{d}{d x} [- 4 y^{2}] + \frac{d}{d x} [3 x^{3}]$

Step 2.2

Evaluate $\frac{d}{d x} [- 4 y^{2}]$ .

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Step 2.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 y^{2}$ with respect to $x$ is $- 4 \frac{d}{d x} [y^{2}]$ .

$0 - 4 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [3 x^{3}]$

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 $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.2.2.1

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

$0 - 4 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [3 x^{3}]$

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 $n = 2$ .

$0 - 4 (2 u \frac{d}{d x} [y]) + \frac{d}{d x} [3 x^{3}]$

Step 2.2.2.3

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

$0 - 4 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [3 x^{3}]$

Step 2.2.3

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

$0 - 4 (2 y y') + \frac{d}{d x} [3 x^{3}]$

Step 2.2.4

Multiply $2$ by $- 4$ .

$0 - 8 y y' + \frac{d}{d x} [3 x^{3}]$

Step 2.3

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

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Step 2.3.1

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

$0 - 8 y y' + 3 \frac{d}{d x} [x^{3}]$

Step 2.3.2

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

$0 - 8 y y' + 3 (3 x^{2})$

Step 2.3.3

Multiply $3$ by $3$ .

$0 - 8 y y' + 9 x^{2}$

Step 2.4

Simplify.

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Step 2.4.1

Subtract $8 y y'$ from $0$ .

$- 8 y y' + 9 x^{2}$

Step 2.4.2

Reorder terms.

$9 x^{2} - 8 y y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 y^{3}$ with respect to $x$ is $- 5 \frac{d}{d x} [y^{3}]$ .

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

Step 3.3.2

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 3.3.2.1

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

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

Step 3.3.2.2

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

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

Step 3.3.2.3

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

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $3$ by $- 5$ .

$3 y' - 15 y^{2} y'$

Step 3.4

Reorder terms.

$- 15 y^{2} y' + 3 y'$

Step 4

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

$9 x^{2} - 8 y y' = - 15 y^{2} y' + 3 y'$

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.

$- 15 y^{2} y' + 3 y' = 9 x^{2} - 8 y y'$

Step 5.2

Add $8 y y'$ to both sides of the equation.

$- 15 y^{2} y' + 3 y' + 8 y y' = 9 x^{2}$

Step 5.3

Factor $y'$ out of $- 15 y^{2} y' + 3 y' + 8 y y'$ .

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Step 5.3.1

Factor $y'$ out of $- 15 y^{2} y'$ .

$y' (- 15 y^{2}) + 3 y' + 8 y y' = 9 x^{2}$

Step 5.3.2

Factor $y'$ out of $3 y'$ .

$y' (- 15 y^{2}) + y' \cdot 3 + 8 y y' = 9 x^{2}$

Step 5.3.3

Factor $y'$ out of $8 y y'$ .

$y' (- 15 y^{2}) + y' \cdot 3 + y' (8 y) = 9 x^{2}$

Step 5.3.4

Factor $y'$ out of $y' (- 15 y^{2}) + y' \cdot 3$ .

$y' (- 15 y^{2} + 3) + y' (8 y) = 9 x^{2}$

Step 5.3.5

Factor $y'$ out of $y' (- 15 y^{2} + 3) + y' (8 y)$ .

$y' (- 15 y^{2} + 3 + 8 y) = 9 x^{2}$

Step 5.4

Reorder terms.

$y' (- 15 y^{2} + 8 y + 3) = 9 x^{2}$

Step 5.5

Divide each term in $y' (- 15 y^{2} + 8 y + 3) = 9 x^{2}$ by $- 15 y^{2} + 8 y + 3$ and simplify.

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Step 5.5.1

Divide each term in $y' (- 15 y^{2} + 8 y + 3) = 9 x^{2}$ by $- 15 y^{2} + 8 y + 3$ .

$\frac{y' (- 15 y^{2} + 8 y + 3)}{- 15 y^{2} + 8 y + 3} = \frac{9 x^{2}}{- 15 y^{2} + 8 y + 3}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $- 15 y^{2} + 8 y + 3$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' (- 15 y^{2} + 8 y + 3)}{- 15 y^{2} + 8 y + 3} = \frac{9 x^{2}}{- 15 y^{2} + 8 y + 3}$

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{9 x^{2}}{- 15 y^{2} + 8 y + 3}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Factor $- 1$ out of $- 15 y^{2}$ .

$y' = \frac{9 x^{2}}{- (15 y^{2}) + 8 y + 3}$

Step 5.5.3.2

Factor $- 1$ out of $8 y$ .

$y' = \frac{9 x^{2}}{- (15 y^{2}) - (- 8 y) + 3}$

Step 5.5.3.3

Factor $- 1$ out of $- (15 y^{2}) - (- 8 y)$ .

$y' = \frac{9 x^{2}}{- (15 y^{2} - 8 y) + 3}$

Step 5.5.3.4

Rewrite $3$ as $- 1 (- 3)$ .

$y' = \frac{9 x^{2}}{- (15 y^{2} - 8 y) - 1 (- 3)}$

Step 5.5.3.5

Factor $- 1$ out of $- (15 y^{2} - 8 y) - 1 (- 3)$ .

$y' = \frac{9 x^{2}}{- (15 y^{2} - 8 y - 3)}$

Step 5.5.3.6

Rewrite negatives.

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Step 5.5.3.6.1

Rewrite $- (15 y^{2} - 8 y - 3)$ as $- 1 (15 y^{2} - 8 y - 3)$ .

$y' = \frac{9 x^{2}}{- 1 (15 y^{2} - 8 y - 3)}$

Step 5.5.3.6.2

Move the negative in front of the fraction.

$y' = - \frac{9 x^{2}}{15 y^{2} - 8 y - 3}$

Step 6

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

$\frac{d y}{d x} = - \frac{9 x^{2}}{15 y^{2} - 8 y - 3}$