Calculus Examples

$y = \log_{4} (x^{2} - 9 x)$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log_{4} (x)$ and $g (x) = x^{2} - 9 x$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $x^{2} - 9 x$ .

$\frac{d}{d u} [\log_{4} (u)] \frac{d}{d x} [x^{2} - 9 x]$

Step 1.2

The derivative of $\log_{4} (u)$ with respect to $u$ is $\frac{1}{u \ln (4)}$ .

$\frac{1}{u \ln (4)} \frac{d}{d x} [x^{2} - 9 x]$

Step 1.3

Replace all occurrences of $u$ with $x^{2} - 9 x$ .

$\frac{1}{(x^{2} - 9 x) \ln (4)} \frac{d}{d x} [x^{2} - 9 x]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} - 9 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x]$ .

$\frac{1}{(x^{2} - 9 x) \ln (4)} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x])$

Step 2.2

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

$\frac{1}{(x^{2} - 9 x) \ln (4)} (2 x + \frac{d}{d x} [- 9 x])$

Step 2.3

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$\frac{1}{(x^{2} - 9 x) \ln (4)} (2 x - 9 \frac{d}{d x} [x])$

Step 2.4

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

$\frac{1}{(x^{2} - 9 x) \ln (4)} (2 x - 9 \cdot 1)$

Step 2.5

Multiply $- 9$ by $1$ .

$\frac{1}{(x^{2} - 9 x) \ln (4)} (2 x - 9)$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\frac{1}{x^{2} \ln (4) - 9 x \ln (4)} (2 x - 9)$

Step 3.2

Reorder the factors of $\frac{1}{x^{2} \ln (4) - 9 x \ln (4)} (2 x - 9)$ .

$(2 x - 9) \frac{1}{x^{2} \ln (4) - 9 x \ln (4)}$

Step 3.3

Factor $x \ln (4)$ out of $x^{2} \ln (4) - 9 x \ln (4)$ .

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Step 3.3.1

Factor $x \ln (4)$ out of $x^{2} \ln (4)$ .

$(2 x - 9) \frac{1}{x \ln (4) (x) - 9 x \ln (4)}$

Step 3.3.2

Factor $x \ln (4)$ out of $- 9 x \ln (4)$ .

$(2 x - 9) \frac{1}{x \ln (4) (x) + x \ln (4) (- 9)}$

Step 3.3.3

Factor $x \ln (4)$ out of $x \ln (4) (x) + x \ln (4) (- 9)$ .

$(2 x - 9) \frac{1}{x \ln (4) (x - 9)}$

Step 3.4

Multiply $2 x - 9$ by $\frac{1}{x \ln (4) (x - 9)}$ .

$\frac{2 x - 9}{x \ln (4) (x - 9)}$