Calculus Examples

$\sin (6 x + 3 y) = x y$

Step 1

Differentiate both sides of the equation.

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

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 $f (x) = \sin (x)$ and $g (x) = 6 x + 3 y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $6 x + 3 y$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [6 x + 3 y]$

Step 2.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [6 x + 3 y]$

Step 2.1.3

Replace all occurrences of $u$ with $6 x + 3 y$ .

$\cos (6 x + 3 y) \frac{d}{d x} [6 x + 3 y]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $6 x + 3 y$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [3 y]$ .

$\cos (6 x + 3 y) (\frac{d}{d x} [6 x] + \frac{d}{d x} [3 y])$

Step 2.2.2

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

$\cos (6 x + 3 y) (6 \frac{d}{d x} [x] + \frac{d}{d x} [3 y])$

Step 2.2.3

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

$\cos (6 x + 3 y) (6 \cdot 1 + \frac{d}{d x} [3 y])$

Step 2.2.4

Multiply $6$ by $1$ .

$\cos (6 x + 3 y) (6 + \frac{d}{d x} [3 y])$

Step 2.2.5

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

$\cos (6 x + 3 y) (6 + 3 \frac{d}{d x} [y])$

Step 2.3

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

$\cos (6 x + 3 y) (6 + 3 y')$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 $g (x) = y$ .

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

Step 3.2

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

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

Step 3.3

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

$x y' + y \cdot 1$

Step 3.4

Multiply $y$ by $1$ .

$x y' + y$

Step 4

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

$\cos (6 x + 3 y) (6 + 3 y') = x y' + y$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $\cos (6 x + 3 y) (6 + 3 y')$ .

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Step 5.1.1.1

Rewrite.

$0 + 0 + \cos (6 x + 3 y) (6 + 3 y') = x y' + y$

Step 5.1.1.2

Simplify by adding zeros.

$\cos (6 x + 3 y) (6 + 3 y') = x y' + y$

Step 5.1.1.3

Apply the distributive property.

$\cos (6 x + 3 y) \cdot 6 + \cos (6 x + 3 y) (3 y') = x y' + y$

Step 5.1.1.4

Simplify the expression.

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Step 5.1.1.4.1

Move $6$ to the left of $\cos (6 x + 3 y)$ .

$6 \cdot \cos (6 x + 3 y) + \cos (6 x + 3 y) (3 y') = x y' + y$

Step 5.1.1.4.2

Rewrite using the commutative property of multiplication.

$6 \cdot \cos (6 x + 3 y) + 3 \cos (6 x + 3 y) y' = x y' + y$

Step 5.1.1.4.3

Reorder factors in $6 \cos (6 x + 3 y) + 3 \cos (6 x + 3 y) y'$ .

$6 \cos (6 x + 3 y) + 3 y' \cos (6 x + 3 y) = x y' + y$

Step 5.2

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

$6 \cos (6 x + 3 y) + 3 y' \cos (6 x + 3 y) - x y' = y$

Step 5.3

Subtract $6 \cos (6 x + 3 y)$ from both sides of the equation.

$3 y' \cos (6 x + 3 y) - x y' = y - 6 \cos (6 x + 3 y)$

Step 5.4

Factor $y'$ out of $3 y' \cos (6 x + 3 y) - x y'$ .

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Step 5.4.1

Factor $y'$ out of $3 y' \cos (6 x + 3 y)$ .

$y' (3 \cos (6 x + 3 y)) - x y' = y - 6 \cos (6 x + 3 y)$

Step 5.4.2

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

$y' (3 \cos (6 x + 3 y)) + y' (- x) = y - 6 \cos (6 x + 3 y)$

Step 5.4.3

Factor $y'$ out of $y' (3 \cos (6 x + 3 y)) + y' (- x)$ .

$y' (3 \cos (6 x + 3 y) - x) = y - 6 \cos (6 x + 3 y)$

Step 5.5

Divide each term in $y' (3 \cos (6 x + 3 y) - x) = y - 6 \cos (6 x + 3 y)$ by $3 \cos (6 x + 3 y) - x$ and simplify.

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Step 5.5.1

Divide each term in $y' (3 \cos (6 x + 3 y) - x) = y - 6 \cos (6 x + 3 y)$ by $3 \cos (6 x + 3 y) - x$ .

$\frac{y' (3 \cos (6 x + 3 y) - x)}{3 \cos (6 x + 3 y) - x} = \frac{y}{3 \cos (6 x + 3 y) - x} + \frac{- 6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $3 \cos (6 x + 3 y) - x$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' (3 \cos (6 x + 3 y) - x)}{3 \cos (6 x + 3 y) - x} = \frac{y}{3 \cos (6 x + 3 y) - x} + \frac{- 6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{y}{3 \cos (6 x + 3 y) - x} + \frac{- 6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Move the negative in front of the fraction.

$y' = \frac{y}{3 \cos (6 x + 3 y) - x} - \frac{6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$

Step 5.5.3.2

Combine the numerators over the common denominator.

$y' = \frac{y - 6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$

Step 6

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

$\frac{d y}{d x} = \frac{y - 6 \cos (6 x + 3 y)}{3 \cos (6 x + 3 y) - x}$