Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $2$ by $- 1$ .

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

Step 2.3

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

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Step 2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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

Step 2.3.1.3

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

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

Step 2.3.2

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

$- 2 x + 2 y y'$

Step 2.4

Reorder terms.

$2 y y' - 2 x$

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 $- 3 y$ with respect to $x$ is $- 3 \frac{d}{d x} [y]$ .

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

Step 3.2

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

$- 3 y'$

Step 4

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

$2 y y' - 2 x = - 3 y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $3 y'$ to both sides of the equation.

$2 y y' - 2 x + 3 y' = 0$

Step 5.2

Add $2 x$ to both sides of the equation.

$2 y y' + 3 y' = 2 x$

Step 5.3

Factor $y'$ out of $2 y y' + 3 y'$ .

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Step 5.3.1

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

$y' (2 y) + 3 y' = 2 x$

Step 5.3.2

Factor $y'$ out of $3 y'$ .

$y' (2 y) + y' \cdot 3 = 2 x$

Step 5.3.3

Factor $y'$ out of $y' (2 y) + y' \cdot 3$ .

$y' (2 y + 3) = 2 x$

Step 5.4

Divide each term in $y' (2 y + 3) = 2 x$ by $2 y + 3$ and simplify.

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Step 5.4.1

Divide each term in $y' (2 y + 3) = 2 x$ by $2 y + 3$ .

$\frac{y' (2 y + 3)}{2 y + 3} = \frac{2 x}{2 y + 3}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $2 y + 3$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (2 y + 3)}{2 y + 3} = \frac{2 x}{2 y + 3}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{2 x}{2 y + 3}$

Step 6

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

$\frac{d y}{d x} = \frac{2 x}{2 y + 3}$