Calculus Examples

$f (x) = \ln (e^{x^{2} - 7 x + 5})$

Step 1

Use logarithm rules to move $x^{2} - 7 x + 5$ out of the exponent.

$\frac{d}{d x} [(x^{2} - 7 x + 5) \ln (e)]$

Step 2

The natural logarithm of $e$ is $1$ .

$\frac{d}{d x} [(x^{2} - 7 x + 5) \cdot 1]$

Step 3

Multiply $x^{2} - 7 x + 5$ by $1$ .

$\frac{d}{d x} [x^{2} - 7 x + 5]$

Step 4

By the Sum Rule, the derivative of $x^{2} - 7 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [5]$

Step 5

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

$2 x + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [5]$

Step 6

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7 x$ with respect to $x$ is $- 7 \frac{d}{d x} [x]$ .

$2 x - 7 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 7

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

$2 x - 7 \cdot 1 + \frac{d}{d x} [5]$

Step 8

Multiply $- 7$ by $1$ .

$2 x - 7 + \frac{d}{d x} [5]$

Step 9

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$2 x - 7 + 0$

Step 10

Add $2 x - 7$ and $0$ .

$2 x - 7$