Calculus Examples

${(\sin (π x) + \cos (π y))}^{2} = 2$

Step 1

Differentiate both sides of the equation.

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

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 $g (x) = \sin (π x) + \cos (π y)$ .

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Step 2.1.1

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

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

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 $n = 2$ .

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

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 $g (x) = π x$ .

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Differentiate.

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Step 2.4.1

Since $π$ is constant with respect to $x$ , the derivative of $π x$ with respect to $x$ is $π \frac{d}{d x} [x]$ .

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

Step 2.4.2

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

$2 (\sin (π x) + \cos (π y)) (\cos (π x) (π \cdot 1) + \frac{d}{d x} [\cos (π y)])$

Step 2.4.3

Multiply $π$ by $1$ .

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

Step 2.5

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 $g (x) = π y$ .

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Step 2.5.1

To apply the Chain Rule, set $u_{3}$ as $π y$ .

$2 (\sin (π x) + \cos (π y)) (\cos (π x) π + \frac{d}{d u_{3}} [\cos (u_{3})] \frac{d}{d x} [π y])$

Step 2.5.2

The derivative of $\cos (u_{3})$ with respect to $u_{3}$ is $- \sin (u_{3})$ .

$2 (\sin (π x) + \cos (π y)) (\cos (π x) π - \sin (u_{3}) \frac{d}{d x} [π y])$

Step 2.5.3

Replace all occurrences of $u_{3}$ with $π y$ .

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

Step 2.6

Since $π$ is constant with respect to $x$ , the derivative of $π y$ with respect to $x$ is $π \frac{d}{d x} [y]$ .

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

Step 2.7

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

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

Step 2.8

Apply the distributive property.

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

Step 3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0$

Step 4

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

$(2 \sin (π x) + 2 \cos (π y)) (\cos (π x) π - \sin (π y) (π y')) = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Divide each term in $(2 \sin (π x) + 2 \cos (π y)) (\cos (π x) π - \sin (π y) π y') = 0$ by $2 \sin (π x) + 2 \cos (π y)$ and simplify.

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Step 5.1.1

Divide each term in $(2 \sin (π x) + 2 \cos (π y)) (\cos (π x) π - \sin (π y) π y') = 0$ by $2 \sin (π x) + 2 \cos (π y)$ .

$\frac{(2 \sin (π x) + 2 \cos (π y)) (\cos (π x) π - \sin (π y) π y')}{2 \sin (π x) + 2 \cos (π y)} = \frac{0}{2 \sin (π x) + 2 \cos (π y)}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $2 \sin (π x) + 2 \cos (π y)$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{(2 \sin (π x) + 2 \cos (π y)) (\cos (π x) π - \sin (π y) π y')}{2 \sin (π x) + 2 \cos (π y)} = \frac{0}{2 \sin (π x) + 2 \cos (π y)}$

Step 5.1.2.1.2

Divide $\cos (π x) π - \sin (π y) π y'$ by $1$ .

$\cos (π x) π - \sin (π y) π y' = \frac{0}{2 \sin (π x) + 2 \cos (π y)}$

Step 5.1.2.2

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

$π \cos (π x) - π y' \sin (π y) = \frac{0}{2 \sin (π x) + 2 \cos (π y)}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Cancel the common factor of $0$ and $2 \sin (π x) + 2 \cos (π y)$ .

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Step 5.1.3.1.1

Factor $2$ out of $0$ .

$π \cos (π x) - π y' \sin (π y) = \frac{2 (0)}{2 \sin (π x) + 2 \cos (π y)}$

Step 5.1.3.1.2

Cancel the common factors.

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Step 5.1.3.1.2.1

Factor $2$ out of $2 \sin (π x)$ .

$π \cos (π x) - π y' \sin (π y) = \frac{2 (0)}{2 (\sin (π x)) + 2 \cos (π y)}$

Step 5.1.3.1.2.2

Factor $2$ out of $2 \cos (π y)$ .

$π \cos (π x) - π y' \sin (π y) = \frac{2 (0)}{2 (\sin (π x)) + 2 (\cos (π y))}$

Step 5.1.3.1.2.3

Factor $2$ out of $2 (\sin (π x)) + 2 (\cos (π y))$ .

$π \cos (π x) - π y' \sin (π y) = \frac{2 (0)}{2 (\sin (π x) + \cos (π y))}$

Step 5.1.3.1.2.4

Cancel the common factor.

$π \cos (π x) - π y' \sin (π y) = \frac{2 \cdot 0}{2 (\sin (π x) + \cos (π y))}$

Step 5.1.3.1.2.5

Rewrite the expression.

$π \cos (π x) - π y' \sin (π y) = \frac{0}{\sin (π x) + \cos (π y)}$

Step 5.1.3.2

Divide $0$ by $\sin (π x) + \cos (π y)$ .

$π \cos (π x) - π y' \sin (π y) = 0$

Step 5.2

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

$- π y' \sin (π y) = - π \cos (π x)$

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Dividing two negative values results in a positive value.

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

Step 5.3.2.2

Cancel the common factor of $π$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Rewrite the expression.

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

Step 5.3.2.3

Cancel the common factor of $\sin (π y)$ .

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Step 5.3.2.3.1

Cancel the common factor.

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

Step 5.3.2.3.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Dividing two negative values results in a positive value.

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

Step 5.3.3.2

Cancel the common factor of $π$ .

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Step 5.3.3.2.1

Cancel the common factor.

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

Step 5.3.3.2.2

Rewrite the expression.

$y' = \frac{\cos (π x)}{\sin (π y)}$

Step 5.3.3.3

Rewrite $\frac{\cos (π x)}{\sin (π y)}$ as a product.

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

Step 5.3.3.4

Write $\cos (π x)$ as a fraction with denominator $1$ .

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

Step 5.3.3.5

Simplify.

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Step 5.3.3.5.1

Divide $\cos (π x)$ by $1$ .

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

Step 5.3.3.5.2

Convert from $\frac{1}{\sin (π y)}$ to $\csc (π y)$ .

$y' = \cos (π x) \csc (π y)$

Step 6

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

$\frac{d y}{d x} = \cos (π x) \csc (π y)$