Calculus Examples

$y = \ln (8 x^{2} - 3 x + 4)$

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 $g (x) = 8 x^{2} - 3 x + 4$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $8 x^{2} - 3 x + 4$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [8 x^{2} - 3 x + 4]$

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [8 x^{2} - 3 x + 4]$

Step 1.3

Replace all occurrences of $u$ with $8 x^{2} - 3 x + 4$ .

$\frac{1}{8 x^{2} - 3 x + 4} \frac{d}{d x} [8 x^{2} - 3 x + 4]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $8 x^{2} - 3 x + 4$ with respect to $x$ is $\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$ .

$\frac{1}{8 x^{2} - 3 x + 4} (\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4])$

Step 2.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$\frac{1}{8 x^{2} - 3 x + 4} (8 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4])$

Step 2.3

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

$\frac{1}{8 x^{2} - 3 x + 4} (8 (2 x) + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4])$

Step 2.4

Multiply $2$ by $8$ .

$\frac{1}{8 x^{2} - 3 x + 4} (16 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4])$

Step 2.5

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [4])$

Step 2.6

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

$\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3 \cdot 1 + \frac{d}{d x} [4])$

Step 2.7

Multiply $- 3$ by $1$ .

$\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3 + \frac{d}{d x} [4])$

Step 2.8

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

$\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3 + 0)$

Step 2.9

Add $16 x - 3$ and $0$ .

$\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3)$

Step 3

Simplify.

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Step 3.1

Reorder the factors of $\frac{1}{8 x^{2} - 3 x + 4} (16 x - 3)$ .

$(16 x - 3) \frac{1}{8 x^{2} - 3 x + 4}$

Step 3.2

Multiply $16 x - 3$ by $\frac{1}{8 x^{2} - 3 x + 4}$ .

$\frac{16 x - 3}{8 x^{2} - 3 x + 4}$