Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $\cos (x y)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\cos (x y)]$

Step 2.1.2

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

$2 u_{1} \frac{d}{d x} [\cos (x y)]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $\cos (x y)$ .

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

Step 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 $g (x) = x y$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{2}$ as $x y$ .

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

Step 2.2.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 2.2.3

Replace all occurrences of $u_{2}$ with $x y$ .

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

Step 2.3

Multiply $- 1$ by $2$ .

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

Step 2.4

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 $g (x) = y$ .

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

Step 2.5

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

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

Step 2.6

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

$- 2 \cos (x y) \sin (x y) (x y' + y \cdot 1)$

Step 2.7

Multiply $y$ by $1$ .

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

Step 2.8

Simplify.

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Step 2.8.1

Apply the distributive property.

$- 2 \cos (x y) \sin (x y) (x y') - 2 \cos (x y) \sin (x y) y$

Step 2.8.2

Reorder terms.

$- 2 x \cos (x y) \sin (x y) y' - 2 y \cos (x y) \sin (x y)$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

$\frac{d}{d x} [x] + \frac{d}{d x} [y]$

Step 3.2

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

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

Step 3.3

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

$1 + y'$

Step 4

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

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

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 $- 2 x \cos (x y) \sin (x y) y' - 2 y \cos (x y) \sin (x y)$ .

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

Step 5.2

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

$- 2 x y' \cos (x y) \sin (x y) - 2 y \cos (x y) \sin (x y) - y' = 1$

Step 5.3

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

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Step 5.3.1

Add $2 y \cos (x y) \sin (x y)$ to both sides of the equation.

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

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Add parentheses.

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

Step 5.3.2.2

Reorder $2 y$ and $\cos (x y) \sin (x y)$ .

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

Step 5.3.2.3

Add parentheses.

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

Step 5.3.2.4

Reorder $\cos (x y)$ and $\sin (x y) \cdot 2$ .

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

Step 5.3.2.5

Reorder $\sin (x y)$ and $2$ .

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

Step 5.3.2.6

Apply the sine double-angle identity.

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

Step 5.3.3

Reorder factors in $1 + \sin (2 x y) y$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

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

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

Step 5.4.3

Factor $y'$ out of $y' (- 2 x \cos (x y) \sin (x y)) + y' \cdot - 1$ .

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

Step 5.5

Divide each term in $y' (- 2 x \cos (x y) \sin (x y) - 1) = 1 + y \sin (2 x y)$ by $- 2 x \cos (x y) \sin (x y) - 1$ and simplify.

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Step 5.5.1

Divide each term in $y' (- 2 x \cos (x y) \sin (x y) - 1) = 1 + y \sin (2 x y)$ by $- 2 x \cos (x y) \sin (x y) - 1$ .

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $- 2 x \cos (x y) \sin (x y) - 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Divide $y'$ by $1$ .

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

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Combine the numerators over the common denominator.

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

Step 5.5.3.2

Factor $- 1$ out of $- 2 x \cos (x y) \sin (x y)$ .

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

Step 5.5.3.3

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 5.5.3.4

Factor $- 1$ out of $- (2 x \cos (x y) \sin (x y)) - 1 (1)$ .

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

Step 5.5.3.5

Rewrite negatives.

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Step 5.5.3.5.1

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

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

Step 5.5.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

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