Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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 $g (x) = x + y$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

$\cos (x + y) (1 + y')$

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 $f (x) = y^{2}$ and $g (x) = \cos (x)$ .

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

Step 3.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Reorder terms.

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

Step 4

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

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

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 $\cos (x + y) (1 + y')$ .

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Step 5.1.1.1

Rewrite.

$0 + 0 + \cos (x + y) (1 + y') = - y^{2} \sin (x) + 2 y \cos (x) y'$

Step 5.1.1.2

Simplify by adding zeros.

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

Step 5.1.1.3

Apply the distributive property.

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

Step 5.1.1.4

Simplify the expression.

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Step 5.1.1.4.1

Multiply $\cos (x + y)$ by $1$ .

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

Step 5.1.1.4.2

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

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

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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

Step 5.3

Subtract $2 y y' \cos (x)$ from both sides of the equation.

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

Step 5.4

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

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.5.3

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

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

Step 5.6

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

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Step 5.6.1

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

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

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

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

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Step 5.6.2.1.1

Cancel the common factor.

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

Step 5.6.2.1.2

Divide $y'$ by $1$ .

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

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Simplify each term.

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Step 5.6.3.1.1

Move the negative in front of the fraction.

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

Step 5.6.3.1.2

Move the negative in front of the fraction.

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

Step 5.6.3.2

Simplify terms.

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Step 5.6.3.2.1

Combine the numerators over the common denominator.

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

Step 5.6.3.2.2

Factor $- 1$ out of $- y^{2} \sin (x)$ .

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

Step 5.6.3.2.3

Factor $- 1$ out of $- \cos (x + y)$ .

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

Step 5.6.3.2.4

Factor $- 1$ out of $- (y^{2} \sin (x)) - (\cos (x + y))$ .

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

Step 5.6.3.2.5

Simplify the expression.

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Step 5.6.3.2.5.1

Rewrite $- (y^{2} \sin (x) + \cos (x + y))$ as $- 1 (y^{2} \sin (x) + \cos (x + y))$ .

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

Step 5.6.3.2.5.2

Move the negative in front of the fraction.

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

Step 6

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

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