Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$4 y^{3} y' - 2 y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [x]$

Step 3.3

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

$12 x^{2} + 1$

Step 4

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

$4 y^{3} y' - 2 y' = 12 x^{2} + 1$

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

$2 y' (2 y^{3}) - 2 y' = 12 x^{2} + 1$

Step 5.1.2

Factor $2 y'$ out of $- 2 y'$ .

$2 y' (2 y^{3}) + 2 y' \cdot - 1 = 12 x^{2} + 1$

Step 5.1.3

Factor $2 y'$ out of $2 y' (2 y^{3}) + 2 y' \cdot - 1$ .

$2 y' (2 y^{3} - 1) = 12 x^{2} + 1$

Step 5.2

Divide each term in $2 y' (2 y^{3} - 1) = 12 x^{2} + 1$ by $2 (2 y^{3} - 1)$ and simplify.

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Step 5.2.1

Divide each term in $2 y' (2 y^{3} - 1) = 12 x^{2} + 1$ by $2 (2 y^{3} - 1)$ .

$\frac{2 y' (2 y^{3} - 1)}{2 (2 y^{3} - 1)} = \frac{12 x^{2}}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{2 y' (2 y^{3} - 1)}{2 (2 y^{3} - 1)} = \frac{12 x^{2}}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.2.1.2

Rewrite the expression.

$\frac{y' (2 y^{3} - 1)}{2 y^{3} - 1} = \frac{12 x^{2}}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.2.2

Cancel the common factor of $2 y^{3} - 1$ .

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Step 5.2.2.2.1

Cancel the common factor.

$\frac{y' (2 y^{3} - 1)}{2 y^{3} - 1} = \frac{12 x^{2}}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{12 x^{2}}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Cancel the common factor of $12$ and $2$ .

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Step 5.2.3.1.1

Factor $2$ out of $12 x^{2}$ .

$y' = \frac{2 (6 x^{2})}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.1.2

Cancel the common factors.

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Step 5.2.3.1.2.1

Cancel the common factor.

$y' = \frac{2 (6 x^{2})}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.1.2.2

Rewrite the expression.

$y' = \frac{6 x^{2}}{2 y^{3} - 1} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.2

To write $\frac{6 x^{2}}{2 y^{3} - 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y' = \frac{6 x^{2}}{2 y^{3} - 1} \cdot \frac{2}{2} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.3

Write each expression with a common denominator of $(2 y^{3} - 1) \cdot 2$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.3.3.1

Multiply $\frac{6 x^{2}}{2 y^{3} - 1}$ by $\frac{2}{2}$ .

$y' = \frac{6 x^{2} \cdot 2}{(2 y^{3} - 1) \cdot 2} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.3.2

Reorder the factors of $(2 y^{3} - 1) \cdot 2$ .

$y' = \frac{6 x^{2} \cdot 2}{2 (2 y^{3} - 1)} + \frac{1}{2 (2 y^{3} - 1)}$

Step 5.2.3.4

Combine the numerators over the common denominator.

$y' = \frac{6 x^{2} \cdot 2 + 1}{2 (2 y^{3} - 1)}$

Step 5.2.3.5

Multiply $2$ by $6$ .

$y' = \frac{12 x^{2} + 1}{2 (2 y^{3} - 1)}$

Step 6

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

$\frac{d y}{d x} = \frac{12 x^{2} + 1}{2 (2 y^{3} - 1)}$