Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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 $f (x) = x$ and $g (x) = \cos (y)$ .

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

Step 2.2.2

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 $g (x) = y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.3

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

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Step 2.3.1

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 $f (x) = y$ and $g (x) = \cos (x)$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Reorder terms.

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

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.

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

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

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

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

Step 5.2

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Move the negative in front of the fraction.

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

Step 5.4.3.2

Combine the numerators over the common denominator.

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

Step 5.4.3.3

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

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

Step 5.4.3.4

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

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

Step 5.4.3.5

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

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

Step 5.4.3.6

Rewrite negatives.

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Step 5.4.3.6.1

Rewrite $- (x \sin (y) - \cos (x))$ as $- 1 (x \sin (y) - \cos (x))$ .

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

Step 5.4.3.6.2

Move the negative in front of the fraction.

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

Step 6

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

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