Calculus Examples

$f (x) = (2 x + 1) \cos (x)$

Step 1

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 $f (x) = 2 x + 1$ and $g (x) = \cos (x)$ .

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

Step 2

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

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

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 3.3

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

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

Step 3.4

Multiply $2$ by $1$ .

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

Step 3.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$(2 x + 1) (- \sin (x)) + \cos (x) (2 + 0)$

Step 3.6

Simplify the expression.

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Step 3.6.1

Add $2$ and $0$ .

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

Step 3.6.2

Move $2$ to the left of $\cos (x)$ .

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

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$2 x (- \sin (x)) + 1 (- \sin (x)) + 2 \cos (x)$

Step 4.2

Combine terms.

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Step 4.2.1

Multiply $- 1$ by $2$ .

$- 2 x \sin (x) + 1 (- \sin (x)) + 2 \cos (x)$

Step 4.2.2

Multiply $- 1$ by $1$ .

$- 2 x \sin (x) - 1 \sin (x) + 2 \cos (x)$

Step 4.2.3

Rewrite $- 1 \sin (x)$ as $- \sin (x)$ .

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