Calculus Examples

$15 x = 15 y + 5 y^{3} + 3 y^{5}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (15 x) = \frac{d}{d x} (15 y + 5 y^{3} + 3 y^{5})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x$ with respect to $x$ is $15 \frac{d}{d x} [x]$ .

$15 \frac{d}{d x} [x]$

Step 2.2

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

$15 \cdot 1$

Step 2.3

Multiply $15$ by $1$ .

$15$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $15 y + 5 y^{3} + 3 y^{5}$ with respect to $x$ is $\frac{d}{d x} [15 y] + \frac{d}{d x} [5 y^{3}] + \frac{d}{d x} [3 y^{5}]$ .

$\frac{d}{d x} [15 y] + \frac{d}{d x} [5 y^{3}] + \frac{d}{d x} [3 y^{5}]$

Step 3.2

Evaluate $\frac{d}{d x} [15 y]$ .

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Step 3.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 y$ with respect to $x$ is $15 \frac{d}{d x} [y]$ .

$15 \frac{d}{d x} [y] + \frac{d}{d x} [5 y^{3}] + \frac{d}{d x} [3 y^{5}]$

Step 3.2.2

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

$15 y' + \frac{d}{d x} [5 y^{3}] + \frac{d}{d x} [3 y^{5}]$

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

$15 y' + 5 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [3 y^{5}]$

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 $u_{1}$ as $y$ .

$15 y' + 5 (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [3 y^{5}]$

Step 3.3.2.2

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

$15 y' + 5 (3 {u_{1}}^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [3 y^{5}]$

Step 3.3.2.3

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

$15 y' + 5 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [3 y^{5}]$

Step 3.3.3

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

$15 y' + 5 (3 y^{2} y') + \frac{d}{d x} [3 y^{5}]$

Step 3.3.4

Multiply $3$ by $5$ .

$15 y' + 15 y^{2} y' + \frac{d}{d x} [3 y^{5}]$

Step 3.4

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

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Step 3.4.1

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

$15 y' + 15 y^{2} y' + 3 \frac{d}{d x} [y^{5}]$

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 $f (x) = x^{5}$ and $g (x) = y$ .

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Step 3.4.2.1

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

$15 y' + 15 y^{2} y' + 3 (\frac{d}{d u_{2}} [{u_{2}}^{5}] \frac{d}{d x} [y])$

Step 3.4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 5$ .

$15 y' + 15 y^{2} y' + 3 (5 {u_{2}}^{4} \frac{d}{d x} [y])$

Step 3.4.2.3

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

$15 y' + 15 y^{2} y' + 3 (5 y^{4} \frac{d}{d x} [y])$

Step 3.4.3

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

$15 y' + 15 y^{2} y' + 3 (5 y^{4} y')$

Step 3.4.4

Multiply $5$ by $3$ .

$15 y' + 15 y^{2} y' + 15 y^{4} y'$

Step 3.5

Reorder terms.

$15 y^{2} y' + 15 y^{4} y' + 15 y'$

Step 4

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

$15 = 15 y^{2} y' + 15 y^{4} y' + 15 y'$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $15 y^{2} y' + 15 y^{4} y' + 15 y' = 15$ .

$15 y^{2} y' + 15 y^{4} y' + 15 y' = 15$

Step 5.2

Factor $15 y'$ out of $15 y^{2} y' + 15 y^{4} y' + 15 y'$ .

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Step 5.2.1

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

$15 y' y^{2} + 15 y^{4} y' + 15 y' = 15$

Step 5.2.2

Factor $15 y'$ out of $15 y^{4} y'$ .

$15 y' y^{2} + 15 y' y^{4} + 15 y' = 15$

Step 5.2.3

Factor $15 y'$ out of $15 y'$ .

$15 y' y^{2} + 15 y' y^{4} + 15 y' \cdot 1 = 15$

Step 5.2.4

Factor $15 y'$ out of $15 y' y^{2} + 15 y' y^{4}$ .

$15 y' (y^{2} + y^{4}) + 15 y' \cdot 1 = 15$

Step 5.2.5

Factor $15 y'$ out of $15 y' (y^{2} + y^{4}) + 15 y' \cdot 1$ .

$15 y' (y^{2} + y^{4} + 1) = 15$

Step 5.3

Reorder terms.

$15 y' (y^{4} + y^{2} + 1) = 15$

Step 5.4

Factor.

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Step 5.4.1

Rewrite $y^{4} + y^{2} + 1$ in a factored form.

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Step 5.4.1.1

Rewrite the middle term.

$15 y' (y^{4} + 2 y^{2} \cdot 1 - y^{2} + 1) = 15$

Step 5.4.1.2

Rearrange terms.

$15 y' (y^{4} + 2 y^{2} \cdot 1 + 1 - y^{2}) = 15$

Step 5.4.1.3

Factor first three terms by perfect square rule.

$15 y' ({(y^{2} + 1)}^{2} - y^{2}) = 15$

Step 5.4.1.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y^{2} + 1$ and $b = y$ .

$15 y' ((y^{2} + 1 + y) (y^{2} + 1 - y)) = 15$

Step 5.4.1.5

Simplify.

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Step 5.4.1.5.1

Reorder terms.

$15 y' ((y^{2} + y + 1) (y^{2} + 1 - y)) = 15$

Step 5.4.1.5.2

Reorder terms.

$15 y' ((y^{2} + y + 1) (y^{2} - y + 1)) = 15$

Step 5.4.2

Remove unnecessary parentheses.

$15 y' (y^{2} + y + 1) (y^{2} - y + 1) = 15$

Step 5.5

Divide each term in $15 y' (y^{2} + y + 1) (y^{2} - y + 1) = 15$ by $15 (y^{2} + y + 1) (y^{2} - y + 1)$ and simplify.

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Step 5.5.1

Divide each term in $15 y' (y^{2} + y + 1) (y^{2} - y + 1) = 15$ by $15 (y^{2} + y + 1) (y^{2} - y + 1)$ .

$\frac{15 y' (y^{2} + y + 1) (y^{2} - y + 1)}{15 (y^{2} + y + 1) (y^{2} - y + 1)} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $15$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{15 y' (y^{2} + y + 1) (y^{2} - y + 1)}{15 (y^{2} + y + 1) (y^{2} - y + 1)} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2.1.2

Rewrite the expression.

$\frac{y' (y^{2} + y + 1) (y^{2} - y + 1)}{(y^{2} + y + 1) (y^{2} - y + 1)} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2.2

Cancel the common factor of $y^{2} + y + 1$ .

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Step 5.5.2.2.1

Cancel the common factor.

$\frac{y' (y^{2} + y + 1) (y^{2} - y + 1)}{(y^{2} + y + 1) (y^{2} - y + 1)} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2.2.2

Rewrite the expression.

$\frac{y' (y^{2} - y + 1)}{y^{2} - y + 1} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2.3

Cancel the common factor of $y^{2} - y + 1$ .

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Step 5.5.2.3.1

Cancel the common factor.

$\frac{y' (y^{2} - y + 1)}{y^{2} - y + 1} = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Cancel the common factor of $15$ .

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Step 5.5.3.1.1

Cancel the common factor.

$y' = \frac{15}{15 (y^{2} + y + 1) (y^{2} - y + 1)}$

Step 5.5.3.1.2

Rewrite the expression.

$y' = \frac{1}{(y^{2} + y + 1) (y^{2} - y + 1)}$

Step 6

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

$\frac{d y}{d x} = \frac{1}{(y^{2} + y + 1) (y^{2} - y + 1)}$