Calculus Examples

$\sin (x y) = y$

Step 1

Differentiate both sides of the equation.

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

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 $f (x) = \sin (x)$ 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 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 $g (x) = y$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Multiply $y$ by $1$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

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

Step 2.6.2

Reorder terms.

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

Step 3

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

$y'$

Step 4

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

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

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

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

Step 5.2

Subtract $y'$ from both sides of the equation.

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

Step 5.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Factor $y'$ out of $- y'$ .

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

Step 5.4.3

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

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

Step 5.5

Divide each term in $y' (x \cos (x y) - 1) = - y \cos (x y)$ by $x \cos (x y) - 1$ and simplify.

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Step 5.5.1

Divide each term in $y' (x \cos (x y) - 1) = - y \cos (x y)$ by $x \cos (x y) - 1$ .

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $x \cos (x y) - 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Divide $y'$ by $1$ .

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

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Move the negative in front of the fraction.

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

Step 6

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

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