Calculus Examples

$q = 0.2802 \ln (x) - 2.505$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (q) = \frac{d}{d x} (0.2802 \ln (x) - 2.505)$

Step 2

The derivative of $q$ with respect to $x$ is $q'$ .

$q'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $0.2802 \ln (x) - 2.505$ with respect to $x$ is $\frac{d}{d x} [0.2802 \ln (x)] + \frac{d}{d x} [- 2.505]$ .

$\frac{d}{d x} [0.2802 \ln (x)] + \frac{d}{d x} [- 2.505]$

Step 3.2

Evaluate $\frac{d}{d x} [0.2802 \ln (x)]$ .

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Step 3.2.1

Since $0.2802$ is constant with respect to $x$ , the derivative of $0.2802 \ln (x)$ with respect to $x$ is $0.2802 \frac{d}{d x} [\ln (x)]$ .

$0.2802 \frac{d}{d x} [\ln (x)] + \frac{d}{d x} [- 2.505]$

Step 3.2.2

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

$0.2802 \frac{1}{x} + \frac{d}{d x} [- 2.505]$

Step 3.2.3

Combine $0.2802$ and $\frac{1}{x}$ .

$\frac{0.2802}{x} + \frac{d}{d x} [- 2.505]$

Step 3.3

Differentiate using the Constant Rule.

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Step 3.3.1

Since $- 2.505$ is constant with respect to $x$ , the derivative of $- 2.505$ with respect to $x$ is $0$ .

$\frac{0.2802}{x} + 0$

Step 3.3.2

Add $\frac{0.2802}{x}$ and $0$ .

$\frac{0.2802}{x}$

Step 4

Reform the equation by setting the left side equal to the right side.

$q' = \frac{0.2802}{x}$

Step 5

Replace $q'$ with $\frac{d q}{d x}$ .

$\frac{d q}{d x} = \frac{0.2802}{x}$