Calculus Examples

$y^{5} + x^{2} y^{3} = 1 + x^{4} y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{5} + x^{2} y^{3}) = \frac{d}{d x} (1 + x^{4} y)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = y$ .

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Step 2.2.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{5}] \frac{d}{d x} [y] + \frac{d}{d x} [x^{2} y^{3}]$

Step 2.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 5$ .

$5 {u_{1}}^{4} \frac{d}{d x} [y] + \frac{d}{d x} [x^{2} y^{3}]$

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

$5 y^{4} y' + x^{2} (3 y^{2} y') + y^{3} (2 x)$

Step 2.3.5

Move $3$ to the left of $x^{2}$ .

$5 y^{4} y' + 3 \cdot x^{2} y^{2} y' + y^{3} (2 x)$

Step 2.3.6

Move $2$ to the left of $y^{3}$ .

$5 y^{4} y' + 3 x^{2} y^{2} y' + 2 y^{3} x$

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 $1 + x^{4} y$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{4} y]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [x^{4} y]$

Step 3.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 3.2

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

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Step 3.2.1

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 $g (x) = y$ .

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

Step 3.2.2

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

$0 + x^{4} y' + y \frac{d}{d x} [x^{4}]$

Step 3.2.3

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

$0 + x^{4} y' + y (4 x^{3})$

Step 3.2.4

Move $4$ to the left of $y$ .

$0 + x^{4} y' + 4 y x^{3}$

Step 3.3

Simplify.

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Step 3.3.1

Add $0$ and $x^{4} y' + 4 y x^{3}$ .

$x^{4} y' + 4 y x^{3}$

Step 3.3.2

Reorder terms.

$x^{4} y' + 4 x^{3} y$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $x^{4} y'$ from both sides of the equation.

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

Step 5.2

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

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

Step 5.3

Factor $y'$ out of $5 y^{4} y' + 3 x^{2} y^{2} y' - x^{4} y'$ .

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Step 5.3.1

Factor $y'$ out of $5 y^{4} y'$ .

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.3.4

Factor $y'$ out of $y' (5 y^{4}) + y' (3 x^{2} y^{2})$ .

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

Step 5.3.5

Factor $y'$ out of $y' (5 y^{4} + 3 x^{2} y^{2}) + y' (- x^{4})$ .

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

Step 5.4

Divide each term in $y' (5 y^{4} + 3 x^{2} y^{2} - x^{4}) = 4 x^{3} y - 2 y^{3} x$ by $5 y^{4} + 3 x^{2} y^{2} - x^{4}$ and simplify.

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Step 5.4.1

Divide each term in $y' (5 y^{4} + 3 x^{2} y^{2} - x^{4}) = 4 x^{3} y - 2 y^{3} x$ by $5 y^{4} + 3 x^{2} y^{2} - x^{4}$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $5 y^{4} + 3 x^{2} y^{2} - x^{4}$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

Factor $2 x y$ out of $4 x^{3} y - 2 y^{3} x$ .

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Step 5.4.3.2.1

Factor $2 x y$ out of $4 x^{3} y$ .

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

Step 5.4.3.2.2

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

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

Step 5.4.3.2.3

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

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

Step 6

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

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