# How do you differentiate f(x)=4x*(x-4)*tanx using the product rule?

Aug 3, 2017

$f ' \left(x\right) = 8 \left(x - 2\right) \left(\tan x\right) + 4 x \left(x - 4\right) \left({\sec}^{2} x\right)$

#### Explanation:

$f \left(x\right) = 4 \left(x\right) \left(x - 4\right) \left(\tan x\right)$

The product rule states that $\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(u \cdot v \cdot w\right) = \left(u\right) ' v w + u \left(v\right) ' w + u v \left(w\right) '}$

I'm going to distribute the polynomial terms to take a shortcut:
$f \left(x\right) = 4 \left({x}^{2} - 4 x\right) \left(\tan x\right)$

$f ' \left(x\right) = 4 \left({x}^{2} - 4 x\right) ' \left(\tan x\right) + 4 \left({x}^{2} - 4 x\right) \left(\tan x\right) '$

$f ' \left(x\right) = 4 \left(2 x - 4\right) \left(\tan x\right) + 4 \left({x}^{2} - 4 x\right) \left({\sec}^{2} x\right)$

Simplifying by factoring:

(Note: don't try substituting the trig identity ${\sec}^{2} x = {\tan}^{2} x + 1$. It just complicates the terms.)

$f ' \left(x\right) = 8 \left(x - 2\right) \left(\tan x\right) + 4 x \left(x - 4\right) \left({\sec}^{2} x\right)$

Aug 3, 2017

$f ' \left(x\right) = 8 \left(x - 2\right) \tan x + 4 x \left(x - 4\right) {\sec}^{2} x$

#### Explanation:

We have:

$f \left(x\right) = 4 x \left(x - 4\right) \tan x$

We will apply the Product Rule for Differentiation:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = u v \frac{\mathrm{dw}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}} w + \frac{\mathrm{du}}{\mathrm{dx}} v w$

$\left\{\begin{matrix}\text{Let" & u = 4x & => & (du)/dx = 4 \\ "And" & v =x-4 & => & (dv)/dx = 1 \\ "And} & w = \tan x & \implies & \frac{\mathrm{dw}}{\mathrm{dx}} = {\sec}^{2} x\end{matrix}\right.$

Then:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = \frac{\mathrm{du}}{\mathrm{dx}} v w + u \frac{\mathrm{dv}}{\mathrm{dx}} w + + u v \frac{\mathrm{dw}}{\mathrm{dx}}$

Gives us:

$f ' \left(x\right) = \left(4\right) \left(x - 4\right) \tan x + \left(4 x\right) \left(1\right) \tan x + + 4 x \left(x - 4\right) {\sec}^{2} x$
$\text{ } = 4 \left\{\left(x - 4\right) \tan x + x \tan x + x \left(x - 4\right) {\sec}^{2} x\right\}$
$\text{ } = 4 \left\{2 x \tan x - 4 \tan x + x \left(x - 4\right) {\sec}^{2} x\right\}$
$\text{ } = 4 \left\{\left(2 x - 4\right) \tan x + x \left(x - 4\right) {\sec}^{2} x\right\}$
$\text{ } = 8 \left(x - 2\right) \tan x + 4 x \left(x - 4\right) {\sec}^{2} x$