Calculus Examples

$x \ln (y) + 6 y = 3 x^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x \ln (y) + 6 y) = \frac{d}{d x} (3 x^{3})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x \ln (y) + 6 y$ with respect to $x$ is $\frac{d}{d x} [x \ln (y)] + \frac{d}{d x} [6 y]$ .

$\frac{d}{d x} [x \ln (y)] + \frac{d}{d x} [6 y]$

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

$x \frac{d}{d x} [\ln (y)] + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [6 y]$

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 $u$ as $y$ .

$x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [y]) + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [6 y]$

Step 2.2.2.2

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

$x (\frac{1}{u} \frac{d}{d x} [y]) + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [6 y]$

Step 2.2.2.3

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

$x (\frac{1}{y} \frac{d}{d x} [y]) + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [6 y]$

Step 2.2.3

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

$x (\frac{1}{y} y') + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [6 y]$

Step 2.2.4

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

$x (\frac{1}{y} y') + \ln (y) \cdot 1 + \frac{d}{d x} [6 y]$

Step 2.2.5

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

$x \frac{y'}{y} + \ln (y) \cdot 1 + \frac{d}{d x} [6 y]$

Step 2.2.6

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

$\frac{x y'}{y} + \ln (y) \cdot 1 + \frac{d}{d x} [6 y]$

Step 2.2.7

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

$\frac{x y'}{y} + \ln (y) + \frac{d}{d x} [6 y]$

Step 2.3

Evaluate $\frac{d}{d x} [6 y]$ .

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Step 2.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 y$ with respect to $x$ is $6 \frac{d}{d x} [y]$ .

$\frac{x y'}{y} + \ln (y) + 6 \frac{d}{d x} [y]$

Step 2.3.2

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

$\frac{x y'}{y} + \ln (y) + 6 y'$

Step 2.4

Reorder terms.

$6 y' + \frac{x y'}{y} + \ln (y)$

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

$3 \frac{d}{d x} [x^{3}]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 (3 x^{2})$

Step 3.3

Multiply $3$ by $3$ .

$9 x^{2}$

Step 4

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

$6 y' + \frac{x y'}{y} + \ln (y) = 9 x^{2}$

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.

$1, y, 1, 1$

Step 5.1.2

The LCM of one and any expression is the expression.

$y$

Step 5.2

Multiply each term in $6 y' + \frac{x y'}{y} + \ln (y) = 9 x^{2}$ by $y$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $6 y' + \frac{x y'}{y} + \ln (y) = 9 x^{2}$ by $y$ .

$6 y' y + \frac{x y'}{y} y + \ln (y) y = 9 x^{2} y$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $y$ .

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Step 5.2.2.1.1

Cancel the common factor.

$6 y' y + \frac{x y'}{y} y + \ln (y) y = 9 x^{2} y$

Step 5.2.2.1.2

Rewrite the expression.

$6 y' y + x y' + \ln (y) y = 9 x^{2} y$

Step 5.2.2.2

Reorder factors in $6 y' y + x y' + \ln (y) y$ .

$6 y' y + x y' + y \ln (y) = 9 x^{2} y$

Step 5.3

Solve the equation.

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Step 5.3.1

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

$6 y' y + x y' = 9 x^{2} y - y \ln (y)$

Step 5.3.2

Factor $y'$ out of $6 y' y + x y'$ .

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Step 5.3.2.1

Factor $y'$ out of $6 y' y$ .

$y' (6 y) + x (y') = 9 x^{2} y - y \ln (y)$

Step 5.3.2.2

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

$y' (6 y) + y' (x) = 9 x^{2} y - y \ln (y)$

Step 5.3.2.3

Factor $y'$ out of $y' (6 y) + y' (x)$ .

$y' (6 y + x) = 9 x^{2} y - y \ln (y)$

Step 5.3.3

Divide each term in $y' (6 y + x) = 9 x^{2} y - y \ln (y)$ by $6 y + x$ and simplify.

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Step 5.3.3.1

Divide each term in $y' (6 y + x) = 9 x^{2} y - y \ln (y)$ by $6 y + x$ .

$\frac{y' (6 y + x)}{6 y + x} = \frac{9 x^{2} y}{6 y + x} + \frac{- y \ln (y)}{6 y + x}$

Step 5.3.3.2

Simplify the left side.

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Step 5.3.3.2.1

Cancel the common factor of $6 y + x$ .

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Step 5.3.3.2.1.1

Cancel the common factor.

$\frac{y' (6 y + x)}{6 y + x} = \frac{9 x^{2} y}{6 y + x} + \frac{- y \ln (y)}{6 y + x}$

Step 5.3.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{9 x^{2} y}{6 y + x} + \frac{- y \ln (y)}{6 y + x}$

Step 5.3.3.3

Simplify the right side.

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Step 5.3.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{9 x^{2} y - y \ln (y)}{6 y + x}$

Step 5.3.3.3.2

Simplify the numerator.

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Step 5.3.3.3.2.1

Factor $y$ out of $9 x^{2} y - y \ln (y)$ .

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Step 5.3.3.3.2.1.1

Factor $y$ out of $9 x^{2} y$ .

$y' = \frac{y (9 x^{2}) - y \ln (y)}{6 y + x}$

Step 5.3.3.3.2.1.2

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

$y' = \frac{y (9 x^{2}) + y (- 1 \ln (y))}{6 y + x}$

Step 5.3.3.3.2.1.3

Factor $y$ out of $y (9 x^{2}) + y (- 1 \ln (y))$ .

$y' = \frac{y (9 x^{2} - 1 \ln (y))}{6 y + x}$

Step 5.3.3.3.2.2

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

$y' = \frac{y (9 x^{2} - \ln (y))}{6 y + x}$

Step 6

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

$\frac{d y}{d x} = \frac{y (9 x^{2} - \ln (y))}{6 y + x}$