Calculus Examples

$5 x^{2} y = - 8 - 8 x y^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (5 x^{2} y) = \frac{d}{d x} (- 8 - 8 x y^{2})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

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

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

Step 2.3

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

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

Step 2.4

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

$5 (x^{2} y' + y (2 x))$

Step 2.5

Move $2$ to the left of $y$ .

$5 (x^{2} y' + 2 \cdot y x)$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$5 (x^{2} y') + 5 (2 y x)$

Step 2.6.2

Multiply $2$ by $5$ .

$5 x^{2} y' + 10 y x$

Step 2.6.3

Reorder terms.

$5 x^{2} y' + 10 x y$

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 $- 8 - 8 x y^{2}$ with respect to $x$ is $\frac{d}{d x} [- 8] + \frac{d}{d x} [- 8 x y^{2}]$ .

$\frac{d}{d x} [- 8] + \frac{d}{d x} [- 8 x y^{2}]$

Step 3.1.2

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 8 x y^{2}]$

Step 3.2

Evaluate $\frac{d}{d x} [- 8 x y^{2}]$ .

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Step 3.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x y^{2}$ with respect to $x$ is $- 8 \frac{d}{d x} [x y^{2}]$ .

$0 - 8 \frac{d}{d x} [x y^{2}]$

Step 3.2.2

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

$0 - 8 (x \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x])$

Step 3.2.3

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 3.2.3.1

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

$0 - 8 (x (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x])$

Step 3.2.3.2

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

$0 - 8 (x (2 u \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x])$

Step 3.2.3.3

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

$0 - 8 (x (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x])$

Step 3.2.4

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

$0 - 8 (x (2 y y') + y^{2} \frac{d}{d x} [x])$

Step 3.2.5

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

$0 - 8 (x (2 y y') + y^{2} \cdot 1)$

Step 3.2.6

Move $2$ to the left of $x$ .

$0 - 8 (2 \cdot x y y' + y^{2} \cdot 1)$

Step 3.2.7

Multiply $y^{2}$ by $1$ .

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

Step 3.3

Simplify.

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Step 3.3.1

Apply the distributive property.

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

Step 3.3.2

Combine terms.

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Step 3.3.2.1

Multiply $2$ by $- 8$ .

$0 - 16 (x y y') - 8 y^{2}$

Step 3.3.2.2

Subtract $- (- 16 x y y' - 8 y^{2})$ from $0$ .

$- 16 x y y' - 8 y^{2}$

Step 3.3.3

Reorder terms.

$- 8 y^{2} - 16 x y y'$

Step 4

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

$5 x^{2} y' + 10 x y = - 8 y^{2} - 16 x y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $16 x y y'$ to both sides of the equation.

$5 x^{2} y' + 10 x y + 16 x y y' = - 8 y^{2}$

Step 5.2

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

$5 x^{2} y' + 16 x y y' = - 8 y^{2} - 10 x y$

Step 5.3

Factor $x y'$ out of $5 x^{2} y' + 16 x y y'$ .

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Step 5.3.1

Factor $x y'$ out of $5 x^{2} y'$ .

$x y' (5 x) + 16 x y y' = - 8 y^{2} - 10 x y$

Step 5.3.2

Factor $x y'$ out of $16 x y y'$ .

$x y' (5 x) + x y' (16 y) = - 8 y^{2} - 10 x y$

Step 5.3.3

Factor $x y'$ out of $x y' (5 x) + x y' (16 y)$ .

$x y' (5 x + 16 y) = - 8 y^{2} - 10 x y$

Step 5.4

Divide each term in $x y' (5 x + 16 y) = - 8 y^{2} - 10 x y$ by $x (5 x + 16 y)$ and simplify.

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Step 5.4.1

Divide each term in $x y' (5 x + 16 y) = - 8 y^{2} - 10 x y$ by $x (5 x + 16 y)$ .

$\frac{x y' (5 x + 16 y)}{x (5 x + 16 y)} = \frac{- 8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $x$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{x y' (5 x + 16 y)}{x (5 x + 16 y)} = \frac{- 8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (5 x + 16 y)}{5 x + 16 y} = \frac{- 8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.2.2

Cancel the common factor of $5 x + 16 y$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (5 x + 16 y)}{5 x + 16 y} = \frac{- 8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.3.1.2

Cancel the common factor of $x$ .

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Step 5.4.3.1.2.1

Cancel the common factor.

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 x y}{x (5 x + 16 y)}$

Step 5.4.3.1.2.2

Rewrite the expression.

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} + \frac{- 10 y}{5 x + 16 y}$

Step 5.4.3.1.3

Move the negative in front of the fraction.

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} - \frac{10 y}{5 x + 16 y}$

Step 5.4.3.2

To write $- \frac{10 y}{5 x + 16 y}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} - \frac{10 y}{5 x + 16 y} \cdot \frac{x}{x}$

Step 5.4.3.3

Write each expression with a common denominator of $x (5 x + 16 y)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.4.3.3.1

Multiply $\frac{10 y}{5 x + 16 y}$ by $\frac{x}{x}$ .

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} - \frac{10 y x}{(5 x + 16 y) x}$

Step 5.4.3.3.2

Reorder the factors of $(5 x + 16 y) x$ .

$y' = - \frac{8 y^{2}}{x (5 x + 16 y)} - \frac{10 y x}{x (5 x + 16 y)}$

Step 5.4.3.4

Combine the numerators over the common denominator.

$y' = \frac{- 8 y^{2} - 10 y x}{x (5 x + 16 y)}$

Step 5.4.3.5

Factor $2 y$ out of $- 8 y^{2} - 10 y x$ .

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Step 5.4.3.5.1

Factor $2 y$ out of $- 8 y^{2}$ .

$y' = \frac{2 y (- 4 y) - 10 y x}{x (5 x + 16 y)}$

Step 5.4.3.5.2

Factor $2 y$ out of $- 10 y x$ .

$y' = \frac{2 y (- 4 y) + 2 y (- 5 x)}{x (5 x + 16 y)}$

Step 5.4.3.5.3

Factor $2 y$ out of $2 y (- 4 y) + 2 y (- 5 x)$ .

$y' = \frac{2 y (- 4 y - 5 x)}{x (5 x + 16 y)}$

Step 5.4.3.6

Factor $- 1$ out of $- 4 y$ .

$y' = \frac{2 y (- (4 y) - 5 x)}{x (5 x + 16 y)}$

Step 5.4.3.7

Factor $- 1$ out of $- 5 x$ .

$y' = \frac{2 y (- (4 y) - (5 x))}{x (5 x + 16 y)}$

Step 5.4.3.8

Factor $- 1$ out of $- (4 y) - (5 x)$ .

$y' = \frac{2 y (- (4 y + 5 x))}{x (5 x + 16 y)}$

Step 5.4.3.9

Simplify the expression.

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Step 5.4.3.9.1

Rewrite $- (4 y + 5 x)$ as $- 1 (4 y + 5 x)$ .

$y' = \frac{2 y (- 1 (4 y + 5 x))}{x (5 x + 16 y)}$

Step 5.4.3.9.2

Move the negative in front of the fraction.

$y' = - \frac{(2 y) (4 y + 5 x)}{x (5 x + 16 y)}$

Step 5.4.3.9.3

Reorder factors in $- \frac{(2 y) (4 y + 5 x)}{x (5 x + 16 y)}$ .

$y' = - \frac{2 y (4 y + 5 x)}{x (5 x + 16 y)}$

Step 6

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

$\frac{d y}{d x} = - \frac{2 y (4 y + 5 x)}{x (5 x + 16 y)}$