Question

For f(x) = (3x + 7) In (2x – 1), find its derivative.

Answer 1

`f'(x) = (6x+14)/(2x-1) + 3\cdot\ln(2x-1)`

Explanation:

We can find the derivative of (the slope of the tangent line to) the function:

`f(x) = (3x + 7) \cdot \ln(2x - 1)`

using the following rules:

• the product rule ...
  `\left[\frac{}{} f(x) \cdot g(x) \frac{}{}\right]' = f(x) \cdot g'(x) + g(x)\cdot f'(x)`
• the chain rule ...
  `\left[\frac{}{}f(g(x)\frac{}{}\right]' = f'\left(\frac{}{}g(x)\frac{}{}\right) \cdot g'(x)`

First, we can apply the product rule, identifying the factors (functions) as:

• 1st:
  `3x + 7`
• 2nd:
  `\ln(2x-1)`

We can find their derivatives by identifying the slope of the first factor and applying the chain rule to the second factor, where the function composition is
`f(x) = \ln(x)` and
`g(x) = 2x-1`.

• 1st:
  `(3x+7)' = 3`
• 2nd:
  `\left[\frac{}{}\ln(2x-1)\frac{}{} \right]' = (2)/(2x-1)`

Next, we plug these values into the product rule to construct an unsimplified form of the derivative.

`f'(x) = (3x+7)\left[(2)/(2x-1)\right] + \ln(2x-1)\left[\frac{}{}3\frac{}{}\right]`

Finally, this can be simplified by executing the possible instances of multiplication.

`\boxed{f'(x) = (6x+14)/(2x-1) + 3\cdot\ln(2x-1)}`