Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

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

$2 y y'$

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

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

Step 3.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [x y]$

Step 3.2

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

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Step 3.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y$ .

$0 + x \frac{d}{d x} [y] + y \frac{d}{d x} [x]$

Step 3.2.2

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

$0 + x y' + y \frac{d}{d x} [x]$

Step 3.2.3

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

$0 + x y' + y \cdot 1$

Step 3.2.4

Multiply $y$ by $1$ .

$0 + x y' + y$

Step 3.3

Add $0$ and $x y' + y$ .

$x y' + y$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $x y'$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y'$ out of $y' (2 y) + y' (- x)$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2 y - x$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{y}{2 y - x}$

Step 6

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

$\frac{d y}{d x} = \frac{y}{2 y - x}$