Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

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 $y$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Multiply $\ln (y)$ by $1$ .

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

Step 2.3

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

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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 $f (x) = x^{3}$ and $g (x) = y$ .

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Step 2.3.1.1

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

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

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 $n = 3$ .

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

Step 2.3.1.3

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

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

Step 2.3.2

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

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

Step 2.4

Reorder terms.

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

Step 3

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

$\frac{1}{x}$

Step 4

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

$3 y^{2} y' + \frac{x y'}{y} + \ln (y) = \frac{1}{x}$

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.

$1, y, 1, x$

Step 5.1.2

Since $1, y, 1, x$ 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 $3 y^{2} y' + \frac{x y'}{y} + \ln (y) = \frac{1}{x}$ by $y x$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $3 y^{2} y' + \frac{x y'}{y} + \ln (y) = \frac{1}{x}$ by $y x$ .

$3 y^{2} y' (y x) + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

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 $y^{2}$ by $y$ by adding the exponents.

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Step 5.2.2.1.1.1

Move $y$ .

$3 (y \cdot y^{2}) y' x + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.1.2

Multiply $y$ by $y^{2}$ .

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Step 5.2.2.1.1.2.1

Raise $y$ to the power of $1$ .

$3 (y^{1} y^{2}) y' x + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 y^{1 + 2} y' x + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.1.3

Add $1$ and $2$ .

$3 y^{3} y' x + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.2

Cancel the common factor of $y$ .

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Step 5.2.2.1.2.1

Factor $y$ out of $y x$ .

$3 y^{3} y' x + \frac{x y'}{y} (y (x)) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.2.2

Cancel the common factor.

$3 y^{3} y' x + \frac{x y'}{y} (y x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.2.3

Rewrite the expression.

$3 y^{3} y' x + x y' x + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.3

Raise $x$ to the power of $1$ .

$3 y^{3} y' x + y' (x^{1} x) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.4

Raise $x$ to the power of $1$ .

$3 y^{3} y' x + y' (x^{1} x^{1}) + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 y^{3} y' x + y' x^{1 + 1} + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.1.6

Add $1$ and $1$ .

$3 y^{3} y' x + y' x^{2} + \ln (y) (y x) = \frac{1}{x} (y x)$

Step 5.2.2.2

Reorder factors in $3 y^{3} y' x + y' x^{2} + \ln (y) (y x)$ .

$3 y^{3} y' x + y' x^{2} + y x \ln (y) = \frac{1}{x} (y x)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Cancel the common factor of $x$ .

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Step 5.2.3.1.1

Factor $x$ out of $y x$ .

$3 y^{3} y' x + y' x^{2} + y x \ln (y) = \frac{1}{x} (x y)$

Step 5.2.3.1.2

Cancel the common factor.

$3 y^{3} y' x + y' x^{2} + y x \ln (y) = \frac{1}{x} (x y)$

Step 5.2.3.1.3

Rewrite the expression.

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

Step 5.3

Solve the equation.

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Step 5.3.1

Subtract $y x \ln (y)$ from both sides of the equation.

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

Step 5.3.2

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

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Step 5.3.2.1

Factor $y' x$ out of $3 y^{3} y' x$ .

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

Step 5.3.2.2

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

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

Step 5.3.2.3

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

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

Simplify the left side.

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Step 5.3.3.2.1

Cancel the common factor of $x$ .

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Step 5.3.3.2.1.1

Cancel the common factor.

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

Step 5.3.3.2.1.2

Rewrite the expression.

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

Step 5.3.3.2.2

Cancel the common factor of $3 y^{3} + x$ .

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Step 5.3.3.2.2.1

Cancel the common factor.

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

Step 5.3.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.3.3.3

Simplify the right side.

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Step 5.3.3.3.1

Simplify each term.

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Step 5.3.3.3.1.1

Cancel the common factor of $x$ .

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Step 5.3.3.3.1.1.1

Cancel the common factor.

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

Step 5.3.3.3.1.1.2

Rewrite the expression.

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

Step 5.3.3.3.1.2

Move the negative in front of the fraction.

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

Step 5.3.3.3.2

To write $- \frac{y \ln (y)}{3 y^{3} + x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 5.3.3.3.3

Write each expression with a common denominator of $x (3 y^{3} + x)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.3.3.3.1

Multiply $\frac{y \ln (y)}{3 y^{3} + x}$ by $\frac{x}{x}$ .

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

Step 5.3.3.3.3.2

Reorder the factors of $(3 y^{3} + x) x$ .

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

Step 5.3.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.3.5

Simplify the numerator.

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Step 5.3.3.3.5.1

Factor $y$ out of $y - y \ln (y) x$ .

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Step 5.3.3.3.5.1.1

Raise $y$ to the power of $1$ .

$y' = \frac{y^{1} - y \ln (y) x}{x (3 y^{3} + x)}$

Step 5.3.3.3.5.1.2

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

$y' = \frac{y \cdot 1 - y \ln (y) x}{x (3 y^{3} + x)}$

Step 5.3.3.3.5.1.3

Factor $y$ out of $- y \ln (y) x$ .

$y' = \frac{y \cdot 1 + y (- 1 \ln (y) x)}{x (3 y^{3} + x)}$

Step 5.3.3.3.5.1.4

Factor $y$ out of $y \cdot 1 + y (- 1 \ln (y) x)$ .

$y' = \frac{y (1 - 1 \ln (y) x)}{x (3 y^{3} + x)}$

Step 5.3.3.3.5.2

Rewrite $- 1 \ln (y)$ as $- \ln (y)$ .

$y' = \frac{y (1 - \ln (y) x)}{x (3 y^{3} + x)}$

Step 5.3.3.3.6

Reorder factors in $\frac{y (1 - \ln (y) x)}{x (3 y^{3} + x)}$ .

$y' = \frac{y (1 - x \ln (y))}{x (3 y^{3} + x)}$

Step 6

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

$\frac{d y}{d x} = \frac{y (1 - x \ln (y))}{x (3 y^{3} + x)}$