Calculus Examples

sin (x + y) = 2 x - 2 y

$\sin (x + y) = 2 x - 2 y$

Step 1

Set

sin (x + y)

$\sin (x + y)$ as a function of

x

$x$ .

f (x) = 2 x - 2 y

$f (x) = 2 x - 2 y$

Step 2

Find the derivative.

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Step 2.1

By the Sum Rule, the derivative of

2 x - 2 y

$2 x - 2 y$ with respect to

x

$x$ is

\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 y]

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 y]$ .

\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 y]

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2 y]$

Step 2.2

Evaluate

\frac{d}{d x} [2 x]

$\frac{d}{d x} [2 x]$ .

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Step 2.2.1

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

2 \frac{d}{d x} [x]

$2 \frac{d}{d x} [x]$ .

2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 y]

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2 y]$

Step 2.2.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

2 \cdot 1 + \frac{d}{d x} [- 2 y]

$2 \cdot 1 + \frac{d}{d x} [- 2 y]$

Step 2.2.3

Multiply

2

$2$ by

1

$1$ .

2 + \frac{d}{d x} [- 2 y]

$2 + \frac{d}{d x} [- 2 y]$

2 + \frac{d}{d x} [- 2 y]

$2 + \frac{d}{d x} [- 2 y]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since

- 2 y

$- 2 y$ is constant with respect to

x

$x$ , the derivative of

- 2 y

$- 2 y$ with respect to

x

$x$ is

0

$0$ .

2 + 0

$2 + 0$

Step 2.3.2

Add

2

$2$ and

0

$0$ .

2

$2$

2

$2$

2

$2$

Step 3

Since

2 \neq 0

$2 \neq 0$ , there are no solutions.

No solution

Step 4

There are no solution found by setting the derivative equal to

0

$0$ ,

2 = 0

$2 = 0$ so there are no horizontal tangent lines.

No horizontal tangent lines found

Step 5