Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply $2$ by $- 1$ .

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

Multiply $- 2$ by $2$ .

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

Reorder terms.

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

Step 3

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

$0$

Step 4

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

$- 4 \sin (2 y) y' + \cos (x) = 0$

Step 5

Solve for $y'$ .

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Simplify the left side.

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

$- 4 y' \sin (2 y) + \cos (x) = 0$

Subtract $\cos (x)$ from both sides of the equation.

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

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

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

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

Simplify the left side.

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Cancel the common factor of $- 4$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $\sin (2 y)$ .

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Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

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Separate fractions.

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

Rewrite $\frac{\cos (x)}{\sin (2 y)}$ as a product.

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

Write $\cos (x)$ as a fraction with denominator $1$ .

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

Simplify.

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Divide $\cos (x)$ by $1$ .

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

Convert from $\frac{1}{\sin (2 y)}$ to $\csc (2 y)$ .

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

Dividing two negative values results in a positive value.

$y' = \frac{1}{4} (\cos (x) \csc (2 y))$

Multiply $\frac{1}{4} (\cos (x) \csc (2 y))$ .

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Combine $\cos (x)$ and $\frac{1}{4}$ .

$y' = \frac{\cos (x)}{4} \csc (2 y)$

Combine $\frac{\cos (x)}{4}$ and $\csc (2 y)$ .

$y' = \frac{\cos (x) \csc (2 y)}{4}$

Step 6

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

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