Calculus Examples

$\ln (x y) - y^{2} = 5$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

Tap for more steps...

Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

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

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 $g (x) = y$ .

Tap for more steps...

Step 2.3.2.1

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

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

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 $n = 2$ .

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

Step 2.3.2.3

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

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

Step 2.3.3

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

$\frac{1}{x y} (x y' + y) - (2 y y')$

Step 2.3.4

Multiply $2$ by $- 1$ .

$\frac{1}{x y} (x y' + y) - 2 y y'$

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Apply the distributive property.

$\frac{1}{x y} (x y') + \frac{1}{x y} y - 2 y y'$

Step 2.4.2

Combine terms.

Tap for more steps...

Step 2.4.2.1

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

$\frac{x}{x y} y' + \frac{1}{x y} y - 2 y y'$

Step 2.4.2.2

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

$\frac{x y'}{x y} + \frac{1}{x y} y - 2 y y'$

Step 2.4.2.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 2.4.2.3.1

Cancel the common factor.

$\frac{x y'}{x y} + \frac{1}{x y} y - 2 y y'$

Step 2.4.2.3.2

Rewrite the expression.

$\frac{y'}{y} + \frac{1}{x y} y - 2 y y'$

Step 2.4.2.4

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

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

Step 2.4.2.5

Cancel the common factor of $y$ .

Tap for more steps...

Step 2.4.2.5.1

Cancel the common factor.

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

Step 2.4.2.5.2

Rewrite the expression.

$\frac{y'}{y} + \frac{1}{x} - 2 y y'$

Step 2.4.3

Reorder terms.

$- 2 y y' + \frac{y'}{y} + \frac{1}{x}$

Step 3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$0$

Step 4

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

$- 2 y y' + \frac{y'}{y} + \frac{1}{x} = 0$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Find the LCD of the terms in the equation.

Tap for more steps...

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.

$1, y, x, 1$

Step 5.1.2

Since $1, y, x, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1, 1$ then find LCM for the variable part $y^{1}, x^{1}$ .

Step 5.1.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 5.1.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 5.1.5

The LCM of $1, 1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 5.1.6

The factor for $y^{1}$ is $y$ itself.

$y^{1} = y$

$y$ occurs $1$ time.

Step 5.1.7

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 5.1.8

The LCM of $y^{1}, x^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$y x$

Step 5.2

Multiply each term in $- 2 y y' + \frac{y'}{y} + \frac{1}{x} = 0$ by $y x$ to eliminate the fractions.

Tap for more steps...

Step 5.2.1

Multiply each term in $- 2 y y' + \frac{y'}{y} + \frac{1}{x} = 0$ by $y x$ .

$- 2 y y' (y x) + \frac{y'}{y} (y x) + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Simplify each term.

Tap for more steps...

Step 5.2.2.1.1

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

Tap for more steps...

Step 5.2.2.1.1.1

Move $y$ .

$- 2 (y \cdot y) y' x + \frac{y'}{y} (y x) + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2.1.1.2

Multiply $y$ by $y$ .

$- 2 y^{2} y' x + \frac{y'}{y} (y x) + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2.1.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 5.2.2.1.2.1

Factor $y$ out of $y x$ .

$- 2 y^{2} y' x + \frac{y'}{y} (y (x)) + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2.1.2.2

Cancel the common factor.

$- 2 y^{2} y' x + \frac{y'}{y} (y x) + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2.1.2.3

Rewrite the expression.

$- 2 y^{2} y' x + y' x + \frac{1}{x} (y x) = 0 (y x)$

Step 5.2.2.1.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.2.2.1.3.1

Factor $x$ out of $y x$ .

$- 2 y^{2} y' x + y' x + \frac{1}{x} (x y) = 0 (y x)$

Step 5.2.2.1.3.2

Cancel the common factor.

$- 2 y^{2} y' x + y' x + \frac{1}{x} (x y) = 0 (y x)$

Step 5.2.2.1.3.3

Rewrite the expression.

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

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

Multiply $0 (y x)$ .

Tap for more steps...

Step 5.2.3.1.1

Multiply $y$ by $0$ .

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

Step 5.2.3.1.2

Multiply $0$ by $x$ .

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

Step 5.3

Solve the equation.

Tap for more steps...

Step 5.3.1

Subtract $y$ from both sides of the equation.

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

Step 5.3.2

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

Tap for more steps...

Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

Factor $y' x$ out of $y' x (- 2 y^{2}) + y' x \cdot 1$ .

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

Step 5.3.3

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

Tap for more steps...

Step 5.3.3.1

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

$\frac{y' x (- 2 y^{2} + 1)}{x (- 2 y^{2} + 1)} = \frac{- y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.3.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 5.3.3.2.1.1

Cancel the common factor.

$\frac{y' x (- 2 y^{2} + 1)}{x (- 2 y^{2} + 1)} = \frac{- y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.2.1.2

Rewrite the expression.

$\frac{y' (- 2 y^{2} + 1)}{- 2 y^{2} + 1} = \frac{- y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.2.2

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

Tap for more steps...

Step 5.3.3.2.2.1

Cancel the common factor.

$\frac{y' (- 2 y^{2} + 1)}{- 2 y^{2} + 1} = \frac{- y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.3.1

Move the negative in front of the fraction.

$y' = - \frac{y}{x (- 2 y^{2} + 1)}$

Step 5.3.3.3.2

Factor $- 1$ out of $- 2 y^{2}$ .

$y' = - \frac{y}{x (- (2 y^{2}) + 1)}$

Step 5.3.3.3.3

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

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

Step 5.3.3.3.4

Factor $- 1$ out of $- (2 y^{2}) - 1 (- 1)$ .

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

Step 5.3.3.3.5

Simplify the expression.

Tap for more steps...

Step 5.3.3.3.5.1

Rewrite $- (2 y^{2} - 1)$ as $- 1 (2 y^{2} - 1)$ .

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

Step 5.3.3.3.5.2

Move the negative in front of the fraction.

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

Step 5.3.3.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.3.5.4

Multiply $\frac{y}{x (2 y^{2} - 1)}$ by $1$ .

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

Step 6

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

$\frac{d y}{d x} = \frac{y}{x (2 y^{2} - 1)}$