Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $- 4 y^{3} - y^{2}$ with respect to $x$ is $\frac{d}{d x} [- 4 y^{3}] + \frac{d}{d x} [- y^{2}]$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $3$ by $- 4$ .

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

Step 2.3

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

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

$- 12 y^{2} y' - \frac{d}{d x} [y^{2}]$

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

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Step 2.3.2.1

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

$- 12 y^{2} y' - (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y])$

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 $n = 2$ .

$- 12 y^{2} y' - (2 u_{2} \frac{d}{d x} [y])$

Step 2.3.2.3

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

$- 12 y^{2} y' - (2 y \frac{d}{d x} [y])$

Step 2.3.3

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

$- 12 y^{2} y' - (2 y y')$

Step 2.3.4

Multiply $2$ by $- 1$ .

$- 12 y^{2} y' - 2 y y'$

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 $3 + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [5 x^{2}]$ .

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

Step 3.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [5 x^{2}]$

Step 3.2

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

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Step 3.2.1

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

$0 + 5 \frac{d}{d x} [x^{2}]$

Step 3.2.2

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

$0 + 5 (2 x)$

Step 3.2.3

Multiply $2$ by $5$ .

$0 + 10 x$

Step 3.3

Add $0$ and $10 x$ .

$10 x$

Step 4

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

$- 12 y^{2} y' - 2 y y' = 10 x$

Step 5

Solve for $y'$ .

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Step 5.1

Factor $2 y y'$ out of $- 12 y^{2} y' - 2 y y'$ .

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Step 5.1.1

Factor $2 y y'$ out of $- 12 y^{2} y'$ .

$2 y y' (- 6 y) - 2 y y' = 10 x$

Step 5.1.2

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

$2 y y' (- 6 y) + 2 y y' \cdot - 1 = 10 x$

Step 5.1.3

Factor $2 y y'$ out of $2 y y' (- 6 y) + 2 y y' \cdot - 1$ .

$2 y y' (- 6 y - 1) = 10 x$

Step 5.2

Divide each term in $2 y y' (- 6 y - 1) = 10 x$ by $2 y (- 6 y - 1)$ and simplify.

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Step 5.2.1

Divide each term in $2 y y' (- 6 y - 1) = 10 x$ by $2 y (- 6 y - 1)$ .

$\frac{2 y y' (- 6 y - 1)}{2 y (- 6 y - 1)} = \frac{10 x}{2 y (- 6 y - 1)}$

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.

$\frac{2 y y' (- 6 y - 1)}{2 y (- 6 y - 1)} = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.2.1.2

Rewrite the expression.

$\frac{y y' (- 6 y - 1)}{y (- 6 y - 1)} = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.2.2

Cancel the common factor of $y$ .

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Step 5.2.2.2.1

Cancel the common factor.

$\frac{y y' (- 6 y - 1)}{y (- 6 y - 1)} = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.2.2.2

Rewrite the expression.

$\frac{y' (- 6 y - 1)}{- 6 y - 1} = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.2.3

Cancel the common factor of $- 6 y - 1$ .

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Step 5.2.2.3.1

Cancel the common factor.

$\frac{y' (- 6 y - 1)}{- 6 y - 1} = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{10 x}{2 y (- 6 y - 1)}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Cancel the common factor of $10$ and $2$ .

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Step 5.2.3.1.1

Factor $2$ out of $10 x$ .

$y' = \frac{2 (5 x)}{2 y (- 6 y - 1)}$

Step 5.2.3.1.2

Cancel the common factors.

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Step 5.2.3.1.2.1

Factor $2$ out of $2 y (- 6 y - 1)$ .

$y' = \frac{2 (5 x)}{2 (y (- 6 y - 1))}$

Step 5.2.3.1.2.2

Cancel the common factor.

$y' = \frac{2 (5 x)}{2 (y (- 6 y - 1))}$

Step 5.2.3.1.2.3

Rewrite the expression.

$y' = \frac{5 x}{y (- 6 y - 1)}$

Step 5.2.3.2

Factor $- 1$ out of $- 6 y$ .

$y' = \frac{5 x}{y (- (6 y) - 1)}$

Step 5.2.3.3

Rewrite $- 1$ as $- 1 (1)$ .

$y' = \frac{5 x}{y (- (6 y) - 1 (1))}$

Step 5.2.3.4

Factor $- 1$ out of $- (6 y) - 1 (1)$ .

$y' = \frac{5 x}{y (- (6 y + 1))}$

Step 5.2.3.5

Rewrite negatives.

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Step 5.2.3.5.1

Rewrite $- (6 y + 1)$ as $- 1 (6 y + 1)$ .

$y' = \frac{5 x}{y (- 1 (6 y + 1))}$

Step 5.2.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{5 x}{y (6 y + 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{5 x}{y (6 y + 1)}$