Calculus Examples

$y \cos (2 x - 4 y) = - 2 x y - 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y \cos (2 x - 4 y)) = \frac{d}{d x} (- 2 x y - 4)$

Step 2

Differentiate the left side of the equation.

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Step 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 $f (x) = y$ and $g (x) = \cos (2 x - 4 y)$ .

$y \frac{d}{d x} [\cos (2 x - 4 y)] + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x - 4 y$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $2 x - 4 y$ .

$y (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x - 4 y]) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.2.2

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

$y (- \sin (u) \frac{d}{d x} [2 x - 4 y]) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.2.3

Replace all occurrences of $u$ with $2 x - 4 y$ .

$y (- \sin (2 x - 4 y) \frac{d}{d x} [2 x - 4 y]) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.3

Differentiate.

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Step 2.3.1

By the Sum Rule, the derivative of $2 x - 4 y$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4 y]$ .

$y (- \sin (2 x - 4 y) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4 y])) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.3.2

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

$y (- \sin (2 x - 4 y) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 4 y])) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.3.3

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

$y (- \sin (2 x - 4 y) (2 \cdot 1 + \frac{d}{d x} [- 4 y])) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.3.4

Multiply $2$ by $1$ .

$y (- \sin (2 x - 4 y) (2 + \frac{d}{d x} [- 4 y])) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.3.5

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

$y (- \sin (2 x - 4 y) (2 - 4 \frac{d}{d x} [y])) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.4

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

$y (- \sin (2 x - 4 y) (2 - 4 y')) + \cos (2 x - 4 y) \frac{d}{d x} [y]$

Step 2.5

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

$y (- \sin (2 x - 4 y) (2 - 4 y')) + \cos (2 x - 4 y) y'$

Step 2.6

Reorder terms.

$- y \sin (2 x - 4 y) (2 - 4 y') + \cos (2 x - 4 y) y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $- 2 x y - 4$ with respect to $x$ is $\frac{d}{d x} [- 2 x y] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [- 2 x y] + \frac{d}{d x} [- 4]$

Step 3.2

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

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Step 3.2.1

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

$- 2 \frac{d}{d x} [x y] + \frac{d}{d x} [- 4]$

Step 3.2.2

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 $g (x) = y$ .

$- 2 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [- 4]$

Step 3.2.3

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

$- 2 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [- 4]$

Step 3.2.4

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

$- 2 (x y' + y \cdot 1) + \frac{d}{d x} [- 4]$

Step 3.2.5

Multiply $y$ by $1$ .

$- 2 (x y' + y) + \frac{d}{d x} [- 4]$

Step 3.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$- 2 (x y' + y) + 0$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$- 2 (x y') - 2 y + 0$

Step 3.4.2

Add $- 2 x y' - 2 y$ and $0$ .

$- 2 x y' - 2 y$

Step 4

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

$- y \sin (2 x - 4 y) (2 - 4 y') + \cos (2 x - 4 y) y' = - 2 x y' - 2 y$

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 $- y \sin (2 x - 4 y) (2 - 4 y') + \cos (2 x - 4 y) y'$ .

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Step 5.1.1.1

Simplify each term.

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Step 5.1.1.1.1

Apply the distributive property.

$- y \sin (2 x - 4 y) \cdot 2 - y \sin (2 x - 4 y) (- 4 y') + \cos (2 x - 4 y) y' = - 2 x y' - 2 y$

Step 5.1.1.1.2

Multiply $2$ by $- 1$ .

$- 2 y \sin (2 x - 4 y) - y \sin (2 x - 4 y) (- 4 y') + \cos (2 x - 4 y) y' = - 2 x y' - 2 y$

Step 5.1.1.1.3

Multiply $- 4$ by $- 1$ .

$- 2 y \sin (2 x - 4 y) + 4 y \sin (2 x - 4 y) y' + \cos (2 x - 4 y) y' = - 2 x y' - 2 y$

Step 5.1.1.2

Reorder factors in $- 2 y \sin (2 x - 4 y) + 4 y \sin (2 x - 4 y) y' + \cos (2 x - 4 y) y'$ .

$- 2 y \sin (2 x - 4 y) + 4 y y' \sin (2 x - 4 y) + y' \cos (2 x - 4 y) = - 2 x y' - 2 y$

Step 5.2

Add $2 x y'$ to both sides of the equation.

$- 2 y \sin (2 x - 4 y) + 4 y y' \sin (2 x - 4 y) + y' \cos (2 x - 4 y) + 2 x y' = - 2 y$

Step 5.3

Add $2 y \sin (2 x - 4 y)$ to both sides of the equation.

$4 y y' \sin (2 x - 4 y) + y' \cos (2 x - 4 y) + 2 x y' = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.4

Factor $y'$ out of $4 y y' \sin (2 x - 4 y) + y' \cos (2 x - 4 y) + 2 x y'$ .

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Step 5.4.1

Factor $y'$ out of $4 y y' \sin (2 x - 4 y)$ .

$y' (4 y \sin (2 x - 4 y)) + y' (\cos (2 x - 4 y)) + 2 x y' = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.4.2

Factor $y'$ out of $y' \cos (2 x - 4 y)$ .

$y' (4 y \sin (2 x - 4 y)) + y' (\cos (2 x - 4 y)) + 2 x y' = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.4.3

Factor $y'$ out of $2 x y'$ .

$y' (4 y \sin (2 x - 4 y)) + y' (\cos (2 x - 4 y)) + y' (2 x) = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.4.4

Factor $y'$ out of $y' (4 y \sin (2 x - 4 y)) + y' (\cos (2 x - 4 y))$ .

$y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y)) + y' (2 x) = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.4.5

Factor $y'$ out of $y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y)) + y' (2 x)$ .

$y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x) = - 2 y + 2 y \sin (2 x - 4 y)$

Step 5.5

Divide each term in $y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x) = - 2 y + 2 y \sin (2 x - 4 y)$ by $4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x$ and simplify.

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Step 5.5.1

Divide each term in $y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x) = - 2 y + 2 y \sin (2 x - 4 y)$ by $4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x$ .

$\frac{y' (4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x} = \frac{- 2 y}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x} + \frac{2 y \sin (2 x - 4 y)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x$ .

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Step 5.5.2.1.1

Cancel the common factor.

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 y}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x} + \frac{2 y \sin (2 x - 4 y)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

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.

$y' = - \frac{2 y}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x} + \frac{2 y \sin (2 x - 4 y)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2

Simplify terms.

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Step 5.5.3.2.1

Combine the numerators over the common denominator.

$y' = \frac{- 2 y + 2 y \sin (2 x - 4 y)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.2

Factor $2 y$ out of $- 2 y + 2 y \sin (2 x - 4 y)$ .

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Step 5.5.3.2.2.1

Factor $2 y$ out of $- 2 y$ .

$y' = \frac{2 y (- 1) + 2 y \sin (2 x - 4 y)}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.2.2

Factor $2 y$ out of $2 y \sin (2 x - 4 y)$ .

$y' = \frac{2 y (- 1) + 2 y (\sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.2.3

Factor $2 y$ out of $2 y (- 1) + 2 y (\sin (2 x - 4 y))$ .

$y' = \frac{2 y (- 1 + \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.3

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

$y' = \frac{2 y (- 1 (1) + \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.4

Factor $- 1$ out of $\sin (2 x - 4 y)$ .

$y' = \frac{2 y (- 1 (1) - 1 (- \sin (2 x - 4 y)))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.5

Factor $- 1$ out of $- 1 (1) - 1 (- \sin (2 x - 4 y))$ .

$y' = \frac{2 y (- 1 (1 - \sin (2 x - 4 y)))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.6

Simplify the expression.

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Step 5.5.3.2.6.1

Move the negative in front of the fraction.

$y' = - \frac{(2 y) (1 - \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 5.5.3.2.6.2

Reorder factors in $- \frac{(2 y) (1 - \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$ .

$y' = - \frac{2 y (1 - \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$

Step 6

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

$\frac{d y}{d x} = - \frac{2 y (1 - \sin (2 x - 4 y))}{4 y \sin (2 x - 4 y) + \cos (2 x - 4 y) + 2 x}$