Calculus Examples

$\ln (4 y) = 5 x y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\ln (4 y)) = \frac{d}{d x} (5 x y)$

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 $f (x) = \ln (x)$ and $g (x) = 4 y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $4 y$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [4 y]$

Step 2.1.2

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

$\frac{1}{u} \frac{d}{d x} [4 y]$

Step 2.1.3

Replace all occurrences of $u$ with $4 y$ .

$\frac{1}{4 y} \frac{d}{d x} [4 y]$

Step 2.2

Differentiate using the Constant Multiple Rule.

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $4 \frac{d}{d x} [y]$ .

$\frac{1}{4 y} (4 \frac{d}{d x} [y])$

Step 2.2.2

Simplify terms.

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Step 2.2.2.1

Combine $4$ and $\frac{1}{4 y}$ .

$\frac{4}{4 y} \frac{d}{d x} [y]$

Step 2.2.2.2

Cancel the common factor of $4$ .

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Step 2.2.2.2.1

Cancel the common factor.

$\frac{4}{4 y} \frac{d}{d x} [y]$

Step 2.2.2.2.2

Rewrite the expression.

$\frac{1}{y} \frac{d}{d x} [y]$

Step 2.3

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

$\frac{1}{y} y'$

Step 2.4

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

$\frac{y'}{y}$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x y$ with respect to $x$ is $5 \frac{d}{d x} [x y]$ .

$5 \frac{d}{d x} [x y]$

Step 3.2

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 $g (x) = y$ .

$5 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 3.3

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

$5 (x y' + y \frac{d}{d x} [x])$

Step 3.4

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

$5 (x y' + y \cdot 1)$

Step 3.5

Multiply $y$ by $1$ .

$5 (x y' + y)$

Step 3.6

Apply the distributive property.

$5 x y' + 5 y$

Step 4

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

$\frac{y'}{y} = 5 x y' + 5 y$

Step 5

Solve for $y'$ .

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Step 5.1

Multiply both sides by $y$ .

$\frac{y'}{y} y = (5 x y' + 5 y) y$

Step 5.2

Simplify.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Cancel the common factor of $y$ .

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Step 5.2.1.1.1

Cancel the common factor.

$\frac{y'}{y} y = (5 x y' + 5 y) y$

Step 5.2.1.1.2

Rewrite the expression.

$y' = (5 x y' + 5 y) y$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $(5 x y' + 5 y) y$ .

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Step 5.2.2.1.1

Apply the distributive property.

$y' = 5 x y' y + 5 y \cdot y$

Step 5.2.2.1.2

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

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Step 5.2.2.1.2.1

Move $y$ .

$y' = 5 x y' y + 5 (y \cdot y)$

Step 5.2.2.1.2.2

Multiply $y$ by $y$ .

$y' = 5 x y' y + 5 y^{2}$

Step 5.2.2.1.3

Move $x$ .

$y' = 5 y' x y + 5 y^{2}$

Step 5.3

Solve for $y'$ .

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Step 5.3.1

Subtract $5 y' x y$ from both sides of the equation.

$y' - 5 y' x y = 5 y^{2}$

Step 5.3.2

Factor $y'$ out of $y' - 5 y' x y$ .

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Step 5.3.2.1

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

$y' \cdot 1 - 5 y' x y = 5 y^{2}$

Step 5.3.2.2

Factor $y'$ out of $- 5 y' x y$ .

$y' \cdot 1 + y' (- 5 x y) = 5 y^{2}$

Step 5.3.2.3

Factor $y'$ out of $y' \cdot 1 + y' (- 5 x y)$ .

$y' (1 - 5 x y) = 5 y^{2}$

Step 5.3.3

Divide each term in $y' (1 - 5 x y) = 5 y^{2}$ by $1 - 5 x y$ and simplify.

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Step 5.3.3.1

Divide each term in $y' (1 - 5 x y) = 5 y^{2}$ by $1 - 5 x y$ .

$\frac{y' (1 - 5 x y)}{1 - 5 x y} = \frac{5 y^{2}}{1 - 5 x y}$

Step 5.3.3.2

Simplify the left side.

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Step 5.3.3.2.1

Cancel the common factor of $1 - 5 x y$ .

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Step 5.3.3.2.1.1

Cancel the common factor.

$\frac{y' (1 - 5 x y)}{1 - 5 x y} = \frac{5 y^{2}}{1 - 5 x y}$

Step 5.3.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{5 y^{2}}{1 - 5 x y}$

Step 6

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

$\frac{d y}{d x} = \frac{5 y^{2}}{1 - 5 x y}$