Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

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 $g (x) = y$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Multiply $y$ by $1$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

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

Step 2.6.2

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $1 + \sin (y)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\sin (y)]$ .

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

Step 3.1.2

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

$0 + \frac{d}{d x} [\sin (y)]$

Step 3.2

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

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Step 3.2.1

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 $g (x) = y$ .

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Step 3.2.1.1

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

$0 + \frac{d}{d u} [\sin (u)] \frac{d}{d x} [y]$

Step 3.2.1.2

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

$0 + \cos (u) \frac{d}{d x} [y]$

Step 3.2.1.3

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

$0 + \cos (y) \frac{d}{d x} [y]$

Step 3.2.2

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

$0 + \cos (y) y'$

Step 3.3

Add $0$ and $\cos (y) y'$ .

$\cos (y) y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

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

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

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.5.3

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

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

Step 5.6

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

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

Step 5.7

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

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Step 5.7.1

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

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

Step 5.7.2

Simplify the left side.

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Step 5.7.2.1

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

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Step 5.7.2.1.1

Cancel the common factor.

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

Step 5.7.2.1.2

Divide $y'$ by $1$ .

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

Step 5.7.3

Simplify the right side.

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Step 5.7.3.1

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

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

Step 5.7.3.2

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

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

Step 5.7.3.3

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

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

Step 5.7.3.4

Rewrite negatives.

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Step 5.7.3.4.1

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

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

Step 5.7.3.4.2

Move the negative in front of the fraction.

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

Step 6

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

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