Calculus Examples

$y = \ln (x^{2} - 4 x + 4)$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} - 4 x + 4$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u$ as $x^{2} - 4 x + 4$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} - 4 x + 4]$

Step 1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [x^{2} - 4 x + 4]$

Step 1.3

Replace all occurrences of $u$ with $x^{2} - 4 x + 4$ .

$\frac{1}{x^{2} - 4 x + 4} \frac{d}{d x} [x^{2} - 4 x + 4]$

Step 2

Differentiate.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $x^{2} - 4 x + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4]$ .

$\frac{1}{x^{2} - 4 x + 4} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4])$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{1}{x^{2} - 4 x + 4} (2 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4])$

Step 2.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$\frac{1}{x^{2} - 4 x + 4} (2 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [4])$

Step 2.4

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

$\frac{1}{x^{2} - 4 x + 4} (2 x - 4 \cdot 1 + \frac{d}{d x} [4])$

Step 2.5

Multiply $- 4$ by $1$ .

$\frac{1}{x^{2} - 4 x + 4} (2 x - 4 + \frac{d}{d x} [4])$

Step 2.6

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{1}{x^{2} - 4 x + 4} (2 x - 4 + 0)$

Step 2.7

Add $2 x - 4$ and $0$ .

$\frac{1}{x^{2} - 4 x + 4} (2 x - 4)$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Reorder the factors of $\frac{1}{x^{2} - 4 x + 4} (2 x - 4)$ .

$(2 x - 4) \frac{1}{x^{2} - 4 x + 4}$

Step 3.2

Factor using the perfect square rule.

Tap for more steps...

Step 3.2.1

Rewrite $4$ as $2^{2}$ .

$(2 x - 4) \frac{1}{x^{2} - 4 x + 2^{2}}$

Step 3.2.2

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

$4 x = 2 \cdot x \cdot 2$

Step 3.2.3

Rewrite the polynomial.

$(2 x - 4) \frac{1}{x^{2} - 2 \cdot x \cdot 2 + 2^{2}}$

Step 3.2.4

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

$(2 x - 4) \frac{1}{{(x - 2)}^{2}}$

Step 3.3

Multiply $2 x - 4$ by $\frac{1}{{(x - 2)}^{2}}$ .

$\frac{2 x - 4}{{(x - 2)}^{2}}$

Step 3.4

Factor $2$ out of $2 x - 4$ .

Tap for more steps...

Step 3.4.1

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 4}{{(x - 2)}^{2}}$

Step 3.4.2

Factor $2$ out of $- 4$ .

$\frac{2 x + 2 \cdot - 2}{{(x - 2)}^{2}}$

Step 3.4.3

Factor $2$ out of $2 x + 2 \cdot - 2$ .

$\frac{2 (x - 2)}{{(x - 2)}^{2}}$

Step 3.5

Cancel the common factor of $x - 2$ and ${(x - 2)}^{2}$ .

Tap for more steps...

Step 3.5.1

Factor $x - 2$ out of $2 (x - 2)$ .

$\frac{(x - 2) \cdot 2}{{(x - 2)}^{2}}$

Step 3.5.2

Cancel the common factors.

Tap for more steps...

Step 3.5.2.1

Factor $x - 2$ out of ${(x - 2)}^{2}$ .

$\frac{(x - 2) \cdot 2}{(x - 2) (x - 2)}$

Step 3.5.2.2

Cancel the common factor.

$\frac{(x - 2) \cdot 2}{(x - 2) (x - 2)}$

Step 3.5.2.3

Rewrite the expression.

$\frac{2}{x - 2}$