Algebra Examples

$y \sin (2 x) = x \cos (2 y)$

Step 1

Differentiate both sides of the equation.

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

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 $g (x) = \sin (2 x)$ .

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

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Simplify the expression.

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Step 2.3.3.1

Multiply $2$ by $1$ .

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

Step 2.3.3.2

Move $2$ to the left of $y \cos (2 x)$ .

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

Step 2.4

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

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

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 $g (x) = \cos (2 y)$ .

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

Step 3.2

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 $g (x) = 2 y$ .

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Replace all occurrences of $u$ with $2 y$ .

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

Step 3.3

Differentiate using the Constant Multiple Rule.

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Step 3.3.1

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

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

Step 3.3.2

Multiply $2$ by $- 1$ .

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

Step 3.4

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

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

Step 3.5

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

$x (- 2 \sin (2 y) y') + \cos (2 y) \cdot 1$

Step 3.6

Simplify the expression.

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Step 3.6.1

Multiply $\cos (2 y)$ by $1$ .

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

Step 3.6.2

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.2

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.3

Subtract $2 y (1 - 2 \sin^{2} (x))$ from both sides of the equation.

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

Step 5.4

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

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

Step 5.5

Simplify the left side.

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Step 5.5.1

Simplify $\sin (2 x) y' + 2 x \sin (2 y) y'$ .

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Step 5.5.1.1

Simplify each term.

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Step 5.5.1.1.1

Apply the sine double-angle identity.

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

Step 5.5.1.1.2

Apply the sine double-angle identity.

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

Step 5.5.1.1.3

Multiply $2$ by $2$ .

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

Step 5.5.1.2

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

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

Step 5.6

Simplify the right side.

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Step 5.6.1

Simplify each term.

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Step 5.6.1.1

Apply the distributive property.

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

Step 5.6.1.2

Multiply $- 2$ by $1$ .

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

Step 5.6.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.6.1.4

Multiply $- 2$ by $- 2$ .

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

Step 5.7

Solve the equation for $y'$ .

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Step 5.7.1

Simplify the left side.

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Step 5.7.1.1

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

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Step 5.7.1.1.1

Simplify each term.

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Step 5.7.1.1.1.1

Add parentheses.

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

Step 5.7.1.1.1.2

Reorder $2 y'$ and $\sin (x) \cos (x)$ .

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

Step 5.7.1.1.1.3

Reorder $\sin (x) \cos (x)$ and $2$ .

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

Step 5.7.1.1.1.4

Apply the sine double-angle identity.

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

Step 5.7.1.1.2

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

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

Step 5.7.2

Simplify the right side.

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Step 5.7.2.1

Apply the cosine double-angle identity.

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

Step 5.7.3

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.7.4

Simplify the left side.

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Step 5.7.4.1

Simplify each term.

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Step 5.7.4.1.1

Apply the sine double-angle identity.

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

Step 5.7.4.1.2

Rewrite using the commutative property of multiplication.

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

Step 5.7.5

Solve the equation for $y'$ .

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Step 5.7.5.1

Simplify the left side.

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Step 5.7.5.1.1

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

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Step 5.7.5.1.1.1

Simplify each term.

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Step 5.7.5.1.1.1.1

Add parentheses.

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

Step 5.7.5.1.1.1.2

Reorder $2 y'$ and $\sin (x) \cos (x)$ .

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

Step 5.7.5.1.1.1.3

Reorder $\sin (x) \cos (x)$ and $2$ .

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

Step 5.7.5.1.1.1.4

Apply the sine double-angle identity.

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

Step 5.7.5.1.1.2

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

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

Step 5.7.5.2

Simplify the right side.

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Step 5.7.5.2.1

Apply the cosine double-angle identity.

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

Step 5.7.5.3

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.7.5.4

Simplify the left side.

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Step 5.7.5.4.1

Simplify each term.

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Step 5.7.5.4.1.1

Apply the sine double-angle identity.

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

Step 5.7.5.4.1.2

Rewrite using the commutative property of multiplication.

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

Step 5.7.5.5

Solve the equation for $y'$ .

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Step 5.7.5.5.1

Simplify the left side.

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Step 5.7.5.5.1.1

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

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Step 5.7.5.5.1.1.1

Simplify each term.

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Step 5.7.5.5.1.1.1.1

Add parentheses.

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

Step 5.7.5.5.1.1.1.2

Reorder $2 y'$ and $\sin (x) \cos (x)$ .

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

Step 5.7.5.5.1.1.1.3

Reorder $\sin (x) \cos (x)$ and $2$ .

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

Step 5.7.5.5.1.1.1.4

Apply the sine double-angle identity.

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

Step 5.7.5.5.1.1.2

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

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

Step 5.7.5.5.2

Simplify the right side.

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Step 5.7.5.5.2.1

Apply the cosine double-angle identity.

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

Step 5.7.5.5.3

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.7.5.5.4

Simplify the left side.

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Step 5.7.5.5.4.1

Simplify each term.

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Step 5.7.5.5.4.1.1

Apply the sine double-angle identity.

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

Step 5.7.5.5.4.1.2

Rewrite using the commutative property of multiplication.

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

Step 5.7.5.5.5

Solve the equation for $y'$ .

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Step 5.7.5.5.5.1

Simplify the left side.

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Step 5.7.5.5.5.1.1

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

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Step 5.7.5.5.5.1.1.1

Simplify each term.

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Step 5.7.5.5.5.1.1.1.1

Add parentheses.

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

Step 5.7.5.5.5.1.1.1.2

Reorder $2 y'$ and $\sin (x) \cos (x)$ .

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

Step 5.7.5.5.5.1.1.1.3

Reorder $\sin (x) \cos (x)$ and $2$ .

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

Step 5.7.5.5.5.1.1.1.4

Apply the sine double-angle identity.

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

Step 5.7.5.5.5.1.1.2

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

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

Step 5.7.5.5.5.2

Simplify the right side.

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Step 5.7.5.5.5.2.1

Apply the cosine double-angle identity.

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

Step 5.7.5.5.5.3

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 5.7.5.5.5.4

Simplify the left side.

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Step 5.7.5.5.5.4.1

Simplify each term.

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Step 5.7.5.5.5.4.1.1

Apply the sine double-angle identity.

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

Step 5.7.5.5.5.4.1.2

Rewrite using the commutative property of multiplication.

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

Step 5.7.5.5.5.5

Solve the equation for $y'$ .

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Step 5.7.5.5.5.5.1

Simplify the left side.

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Step 5.7.5.5.5.5.1.1

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

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Step 5.7.5.5.5.5.1.1.1

Simplify each term.

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Step 5.7.5.5.5.5.1.1.1.1

Add parentheses.

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

Step 5.7.5.5.5.5.1.1.1.2

Reorder $2 y'$ and $\sin (x) \cos (x)$ .

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

Step 5.7.5.5.5.5.1.1.1.3

Reorder $\sin (x) \cos (x)$ and $2$ .

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

Step 5.7.5.5.5.5.1.1.1.4

Apply the sine double-angle identity.

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

Step 5.7.5.5.5.5.1.1.2

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

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

Step 5.7.5.5.5.5.2

Simplify the right side.

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Step 5.7.5.5.5.5.2.1

Apply the cosine double-angle identity.

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

Step 5.7.5.5.5.5.3

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

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Step 5.7.5.5.5.5.3.1

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

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

Step 5.7.5.5.5.5.3.2

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

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

Step 5.7.5.5.5.5.3.3

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

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

Step 5.7.5.5.5.5.4

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

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Step 5.7.5.5.5.5.4.1

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

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

Step 5.7.5.5.5.5.4.2

Simplify the left side.

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Step 5.7.5.5.5.5.4.2.1

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

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Step 5.7.5.5.5.5.4.2.1.1

Cancel the common factor.

Step 5.7.5.5.5.5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.7.5.5.5.5.4.3

Simplify the right side.

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Step 5.7.5.5.5.5.4.3.1

Move the negative in front of the fraction.

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

Step 5.7.5.5.5.5.4.3.2

Combine into one fraction.

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Step 5.7.5.5.5.5.4.3.2.1

Combine the numerators over the common denominator.

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

Step 5.7.5.5.5.5.4.3.2.2

Combine the numerators over the common denominator.

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

Step 6

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

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