Calculus Examples

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

Step 1

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

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

Step 2

Evaluate $\frac{d}{d x} [\sin (x + y)]$ .

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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 $x + y$ .

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

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

$\cos (x + y) (1 + \frac{d}{d x} [y]) + \frac{d}{d x} [- x y]$

Step 2.4

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

$\cos (x + y) (1 + 0) + \frac{d}{d x} [- x y]$

Step 2.5

Add $1$ and $0$ .

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

Step 2.6

Multiply $\cos (x + y)$ by $1$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Multiply $- 1$ by $1$ .

$\cos (x + y) - y$

Step 4

Reorder terms.

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