Calculus Examples

$x^{2} - 3 \ln (y) + y^{2} = 10$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} - 3 \ln (y) + y^{2}) = \frac{d}{d x} (10)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

Evaluate $\frac{d}{d x} [- 3 \ln (y)]$ .

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Step 2.2.1

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

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

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 $f (x) = \ln (x)$ and $g (x) = y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Combine $\frac{1}{y}$ and $y'$ .

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

Step 2.2.5

Combine $- 3$ and $\frac{y'}{y}$ .

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

Step 2.2.6

Move the negative in front of the fraction.

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

Step 2.3

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

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Step 2.3.1

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 $u_{2}$ as $y$ .

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

Step 2.3.1.2

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

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

Step 2.3.1.3

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

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

Step 2.3.2

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

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

Step 2.4

Reorder terms.

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

Step 3

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$0$

Step 4

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

$2 y y' + 2 x - \frac{3 y'}{y} = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Find the LCD of the terms in the equation.

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Step 5.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 1, y, 1$

Step 5.1.2

The LCM of one and any expression is the expression.

$y$

Step 5.2

Multiply each term in $2 y y' + 2 x - \frac{3 y'}{y} = 0$ by $y$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $2 y y' + 2 x - \frac{3 y'}{y} = 0$ by $y$ .

$2 y y' y + 2 x y - \frac{3 y'}{y} y = 0 y$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply $y$ by $y$ by adding the exponents.

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Step 5.2.2.1.1.1

Move $y$ .

$2 (y \cdot y) y' + 2 x y - \frac{3 y'}{y} y = 0 y$

Step 5.2.2.1.1.2

Multiply $y$ by $y$ .

$2 y^{2} y' + 2 x y - \frac{3 y'}{y} y = 0 y$

Step 5.2.2.1.2

Cancel the common factor of $y$ .

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Step 5.2.2.1.2.1

Move the leading negative in $- \frac{3 y'}{y}$ into the numerator.

$2 y^{2} y' + 2 x y + \frac{- 3 y'}{y} y = 0 y$

Step 5.2.2.1.2.2

Cancel the common factor.

$2 y^{2} y' + 2 x y + \frac{- 3 y'}{y} y = 0 y$

Step 5.2.2.1.2.3

Rewrite the expression.

$2 y^{2} y' + 2 x y - 3 y' = 0 y$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Multiply $0$ by $y$ .

$2 y^{2} y' + 2 x y - 3 y' = 0$

Step 5.3

Solve the equation.

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Step 5.3.1

Subtract $2 x y$ from both sides of the equation.

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

Step 5.3.2

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

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

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

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

Simplify the left side.

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Step 5.3.3.2.1

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

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Step 5.3.3.2.1.1

Cancel the common factor.

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

Step 5.3.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3.3

Simplify the right side.

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Step 5.3.3.3.1

Move the negative in front of the fraction.

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

Step 6

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

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