Algebra Examples

$2 \sin (x) \cos (y) = 1$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = \sin (x)$ and $g (y) = \cos (y)$ .

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Reorder terms.

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

Step 3

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$0$

Step 4

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

$- 2 \sin (x) \sin (y) + 2 \cos (x) \cos (y) x' = 0$

Step 5

Solve for $x'$ .

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Simplify the left side.

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

$- 2 \sin (x) \sin (y) + 2 x' \cos (x) \cos (y) = 0$

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

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

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

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

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $\cos (x)$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $\cos (y)$ .

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Cancel the common factor.

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

Divide $x'$ by $1$ .

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

Simplify the right side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Separate fractions.

$x' = \frac{\sin (y)}{\cos (y)} \cdot \frac{\sin (x)}{\cos (x)}$

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$x' = \frac{\sin (y)}{\cos (y)} \tan (x)$

Convert from $\frac{\sin (y)}{\cos (y)}$ to $\tan (y)$ .

$x' = \tan (y) \tan (x)$

Step 6

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

$\frac{d x}{d y} = \tan (y) \tan (x)$