Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 y^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [y^{2}]$ .

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

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $- 3$ .

$- 6 y y' + \frac{d}{d x} [- y] + \frac{d}{d x} [5 x^{2}]$

Step 2.3

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

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Step 2.3.1

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

$- 6 y y' - \frac{d}{d x} [y] + \frac{d}{d x} [5 x^{2}]$

Step 2.3.2

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

$- 6 y y' - y' + \frac{d}{d x} [5 x^{2}]$

Step 2.4

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

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Step 2.4.1

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

$- 6 y y' - y' + 5 \frac{d}{d x} [x^{2}]$

Step 2.4.2

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

$- 6 y y' - y' + 5 (2 x)$

Step 2.4.3

Multiply $2$ by $5$ .

$- 6 y y' - y' + 10 x$

Step 2.5

Reorder terms.

$- 6 y y' + 10 x - y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 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 3.2.1

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

$- 3 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Multiply $3$ by $- 3$ .

$- 9 (y^{2} \frac{d}{d x} [y])$

Step 3.4

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

$- 9 y^{2} y'$

Step 4

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

$- 6 y y' + 10 x - y' = - 9 y^{2} y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $9 y^{2} y'$ to both sides of the equation.

$- 6 y y' + 10 x - y' + 9 y^{2} y' = 0$

Step 5.2

Subtract $10 x$ from both sides of the equation.

$- 6 y y' - y' + 9 y^{2} y' = - 10 x$

Step 5.3

Factor $y'$ out of $- 6 y y' - y' + 9 y^{2} y'$ .

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Step 5.3.1

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

$y' (- 6 y) - y' + 9 y^{2} y' = - 10 x$

Step 5.3.2

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

$y' (- 6 y) + y' \cdot - 1 + 9 y^{2} y' = - 10 x$

Step 5.3.3

Factor $y'$ out of $9 y^{2} y'$ .

$y' (- 6 y) + y' \cdot - 1 + y' (9 y^{2}) = - 10 x$

Step 5.3.4

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

$y' (- 6 y - 1) + y' (9 y^{2}) = - 10 x$

Step 5.3.5

Factor $y'$ out of $y' (- 6 y - 1) + y' (9 y^{2})$ .

$y' (- 6 y - 1 + 9 y^{2}) = - 10 x$

Step 5.4

Reorder terms.

$y' (9 y^{2} - 6 y - 1) = - 10 x$

Step 5.5

Divide each term in $y' (9 y^{2} - 6 y - 1) = - 10 x$ by $9 y^{2} - 6 y - 1$ and simplify.

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Step 5.5.1

Divide each term in $y' (9 y^{2} - 6 y - 1) = - 10 x$ by $9 y^{2} - 6 y - 1$ .

$\frac{y' (9 y^{2} - 6 y - 1)}{9 y^{2} - 6 y - 1} = \frac{- 10 x}{9 y^{2} - 6 y - 1}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $9 y^{2} - 6 y - 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' (9 y^{2} - 6 y - 1)}{9 y^{2} - 6 y - 1} = \frac{- 10 x}{9 y^{2} - 6 y - 1}$

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 10 x}{9 y^{2} - 6 y - 1}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Move the negative in front of the fraction.

$y' = - \frac{10 x}{9 y^{2} - 6 y - 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{10 x}{9 y^{2} - 6 y - 1}$