Calculus Examples

$\sqrt{x^{2} - 2 x + 1}$

Step 1

Rewrite $\frac{d}{d x} [\sqrt{x^{2} - 2 x + 1}]$ as $\frac{d}{d x} [x - 1]$ .

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Step 1.1

Factor using the perfect square rule.

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Step 1.1.1

Rewrite $1$ as $1^{2}$ .

$\frac{d}{d x} [\sqrt{x^{2} - 2 x + 1^{2}}]$

Step 1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 1.1.3

Rewrite the polynomial.

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

Step 1.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{d}{d x} [\sqrt{{(x - 1)}^{2}}]$

Step 1.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{d}{d x} [x - 1]$

Step 2

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

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

Step 3

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

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

Step 4

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

$1 + 0$

Step 5

Add $1$ and $0$ .

$1$