Calculus Examples

$5 x^{3} = y^{3} - 4 y$

Step 1

Differentiate both sides of the equation.

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

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 $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

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

Step 2.2

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

$5 (3 x^{2})$

Step 2.3

Multiply $3$ by $5$ .

$15 x^{2}$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 4

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

$15 x^{2} = 3 y^{2} y' - 4 y'$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $3 y^{2} y' - 4 y' = 15 x^{2}$ .

$3 y^{2} y' - 4 y' = 15 x^{2}$

Step 5.2

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

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Step 5.2.1

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

$y' (3 y^{2}) - 4 y' = 15 x^{2}$

Step 5.2.2

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

$y' (3 y^{2}) + y' \cdot - 4 = 15 x^{2}$

Step 5.2.3

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

$y' (3 y^{2} - 4) = 15 x^{2}$

Step 5.3

Divide each term in $y' (3 y^{2} - 4) = 15 x^{2}$ by $3 y^{2} - 4$ and simplify.

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Step 5.3.1

Divide each term in $y' (3 y^{2} - 4) = 15 x^{2}$ by $3 y^{2} - 4$ .

$\frac{y' (3 y^{2} - 4)}{3 y^{2} - 4} = \frac{15 x^{2}}{3 y^{2} - 4}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (3 y^{2} - 4)}{3 y^{2} - 4} = \frac{15 x^{2}}{3 y^{2} - 4}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{15 x^{2}}{3 y^{2} - 4}$

Step 6

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

$\frac{d y}{d x} = \frac{15 x^{2}}{3 y^{2} - 4}$