Calculus Examples

$y = \sin (x y)$

Step 1

Differentiate both sides of the equation.

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

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 3.1.1

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

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

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.2

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 3.3

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

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

Step 3.4

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 3.5

Multiply $y$ by $1$ .

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

Step 3.6

Simplify.

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Step 3.6.1

Apply the distributive property.

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

Step 3.6.2

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the right side.

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Step 5.1.1

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

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

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

Factor $y'$ out of ${y'}^{1}$ .

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 6

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

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