# How do I find the derivative of f(x)=(16x^2+1) ln (x-3)?

Dec 22, 2015

This question will demand the use of chain rule and product rule, as well as the knowledge of how to derivate a logarithmic function.

#### Explanation:

First of all, we have a product of two terms:

• $16 {x}^{2} + 1$
• $\ln \left(x - 3\right)$

However, one cannot derivate straightforward a logarithmic function with a function in it. Instead, we must apply chain rule and rename $u = x - 3$ so now we can proceed. The chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{u}\right) \left(1\right) = \frac{1}{u} = \frac{1}{x - 3}$

Fine. Now, we can apply the product rule, which states that, for $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$, $f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$, as follows:

(where the symbol $'$ indicates the first derivative)

$f ' \left(x\right) = 32 x \ln \left(x - 3\right) + \left(16 {x}^{2} + 1\right) \left(\frac{1}{x - 3}\right)$

$f ' \left(x\right) = 32 x \ln \left(x - 3\right) + \frac{16 {x}^{2} + 1}{x - 3}$

The struggle to develop and try to simplify it is not worth it. This is your final answer!