Calculus Examples

$3 x^{4} y^{2} - 7 x y^{3} = 4 - 8 y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 x^{4} y^{2} - 7 x y^{3}) = \frac{d}{d x} (4 - 8 y)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $3 x^{4} y^{2} - 7 x y^{3}$ with respect to $x$ is $\frac{d}{d x} [3 x^{4} y^{2}] + \frac{d}{d x} [- 7 x y^{3}]$ .

$\frac{d}{d x} [3 x^{4} y^{2}] + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2

Evaluate $\frac{d}{d x} [3 x^{4} y^{2}]$ .

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Step 2.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{4} y^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{4} y^{2}]$ .

$3 \frac{d}{d x} [x^{4} y^{2}] + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.2

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 $f (x) = x^{4}$ and $g (x) = y^{2}$ .

$3 (x^{4} \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.3

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 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$3 (x^{4} (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.3.2

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

$3 (x^{4} (2 u_{1} \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $y$ .

$3 (x^{4} (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.4

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

$3 (x^{4} (2 y y') + y^{2} \frac{d}{d x} [x^{4}]) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$3 (x^{4} (2 y y') + y^{2} (4 x^{3})) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.6

Move $2$ to the left of $x^{4}$ .

$3 (2 \cdot x^{4} y y' + y^{2} (4 x^{3})) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.2.7

Move $4$ to the left of $y^{2}$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) + \frac{d}{d x} [- 7 x y^{3}]$

Step 2.3

Evaluate $\frac{d}{d x} [- 7 x y^{3}]$ .

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Step 2.3.1

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7 x y^{3}$ with respect to $x$ is $- 7 \frac{d}{d x} [x y^{3}]$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 \frac{d}{d x} [x y^{3}]$

Step 2.3.2

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 $g (x) = y^{3}$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x \frac{d}{d x} [y^{3}] + y^{3} \frac{d}{d x} [x])$

Step 2.3.3

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 2.3.3.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x (3 {u_{2}}^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.3.3

Replace all occurrences of $u_{2}$ with $y$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x (3 y^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.4

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

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x (3 y^{2} y') + y^{3} \frac{d}{d x} [x])$

Step 2.3.5

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

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (x (3 y^{2} y') + y^{3} \cdot 1)$

Step 2.3.6

Move $3$ to the left of $x$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (3 \cdot x y^{2} y' + y^{3} \cdot 1)$

Step 2.3.7

Multiply $y^{3}$ by $1$ .

$3 (2 x^{4} y y' + 4 y^{2} x^{3}) - 7 (3 x y^{2} y' + y^{3})$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$3 (2 x^{4} y y') + 3 (4 y^{2} x^{3}) - 7 (3 x y^{2} y' + y^{3})$

Step 2.4.2

Apply the distributive property.

$3 (2 x^{4} y y') + 3 (4 y^{2} x^{3}) - 7 (3 x y^{2} y') - 7 y^{3}$

Step 2.4.3

Combine terms.

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Step 2.4.3.1

Multiply $2$ by $3$ .

$6 (x^{4} y y') + 3 (4 y^{2} x^{3}) - 7 (3 x y^{2} y') - 7 y^{3}$

Step 2.4.3.2

Multiply $4$ by $3$ .

$6 x^{4} y y' + 12 (y^{2} x^{3}) - 7 (3 x y^{2} y') - 7 y^{3}$

Step 2.4.3.3

Multiply $3$ by $- 7$ .

$6 x^{4} y y' + 12 y^{2} x^{3} - 21 x y^{2} y' - 7 y^{3}$

Step 2.4.4

Reorder terms.

$- 7 y^{3} + 6 x^{4} y y' + 12 x^{3} y^{2} - 21 y^{2} x y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $4 - 8 y$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 8 y]$ .

$\frac{d}{d x} [4] + \frac{d}{d x} [- 8 y]$

Step 3.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [- 8 y]$ .

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Step 3.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 y$ with respect to $x$ is $- 8 \frac{d}{d x} [y]$ .

$0 - 8 \frac{d}{d x} [y]$

Step 3.2.2

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

$0 - 8 y'$

Step 3.3

Subtract $8 y'$ from $0$ .

$- 8 y'$

Step 4

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

$- 7 y^{3} + 6 x^{4} y y' + 12 x^{3} y^{2} - 21 y^{2} x y' = - 8 y'$

Step 5

Solve for $y'$ .

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Step 5.1

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

$- 7 y^{3} + 6 x^{4} y y' + 12 x^{3} y^{2} - 21 y^{2} x y' + 8 y' = 0$

Step 5.2

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.2.1

Add $7 y^{3}$ to both sides of the equation.

$6 x^{4} y y' + 12 x^{3} y^{2} - 21 y^{2} x y' + 8 y' = 7 y^{3}$

Step 5.2.2

Subtract $12 x^{3} y^{2}$ from both sides of the equation.

$6 x^{4} y y' - 21 y^{2} x y' + 8 y' = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.3

Factor $y'$ out of $6 x^{4} y y' - 21 y^{2} x y' + 8 y'$ .

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Step 5.3.1

Factor $y'$ out of $6 x^{4} y y'$ .

$y' (6 x^{4} y) - 21 y^{2} x y' + 8 y' = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.3.2

Factor $y'$ out of $- 21 y^{2} x y'$ .

$y' (6 x^{4} y) + y' (- 21 y^{2} x) + 8 y' = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.3.3

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

$y' (6 x^{4} y) + y' (- 21 y^{2} x) + y' \cdot 8 = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.3.4

Factor $y'$ out of $y' (6 x^{4} y) + y' (- 21 y^{2} x)$ .

$y' (6 x^{4} y - 21 y^{2} x) + y' \cdot 8 = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.3.5

Factor $y'$ out of $y' (6 x^{4} y - 21 y^{2} x) + y' \cdot 8$ .

$y' (6 x^{4} y - 21 y^{2} x + 8) = 7 y^{3} - 12 x^{3} y^{2}$

Step 5.4

Divide each term in $y' (6 x^{4} y - 21 y^{2} x + 8) = 7 y^{3} - 12 x^{3} y^{2}$ by $6 x^{4} y - 21 y^{2} x + 8$ and simplify.

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Step 5.4.1

Divide each term in $y' (6 x^{4} y - 21 y^{2} x + 8) = 7 y^{3} - 12 x^{3} y^{2}$ by $6 x^{4} y - 21 y^{2} x + 8$ .

$\frac{y' (6 x^{4} y - 21 y^{2} x + 8)}{6 x^{4} y - 21 y^{2} x + 8} = \frac{7 y^{3}}{6 x^{4} y - 21 y^{2} x + 8} + \frac{- 12 x^{3} y^{2}}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $6 x^{4} y - 21 y^{2} x + 8$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (6 x^{4} y - 21 y^{2} x + 8)}{6 x^{4} y - 21 y^{2} x + 8} = \frac{7 y^{3}}{6 x^{4} y - 21 y^{2} x + 8} + \frac{- 12 x^{3} y^{2}}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{7 y^{3}}{6 x^{4} y - 21 y^{2} x + 8} + \frac{- 12 x^{3} y^{2}}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{7 y^{3} - 12 x^{3} y^{2}}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.3.2

Factor $y^{2}$ out of $7 y^{3} - 12 x^{3} y^{2}$ .

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Step 5.4.3.2.1

Factor $y^{2}$ out of $7 y^{3}$ .

$y' = \frac{y^{2} (7 y) - 12 x^{3} y^{2}}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.3.2.2

Factor $y^{2}$ out of $- 12 x^{3} y^{2}$ .

$y' = \frac{y^{2} (7 y) + y^{2} (- 12 x^{3})}{6 x^{4} y - 21 y^{2} x + 8}$

Step 5.4.3.2.3

Factor $y^{2}$ out of $y^{2} (7 y) + y^{2} (- 12 x^{3})$ .

$y' = \frac{y^{2} (7 y - 12 x^{3})}{6 x^{4} y - 21 y^{2} x + 8}$

Step 6

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

$\frac{d y}{d x} = \frac{y^{2} (7 y - 12 x^{3})}{6 x^{4} y - 21 y^{2} x + 8}$