# How do you differentiate f(x)= ( x tanx )/ (x + 4 ) using the quotient rule?

Oct 25, 2017

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

#### Explanation:

If you'd like additional help on the concepts associated with this problem, I have some videos that can help:

Anyways, onto the problem:

For this, you'll actually need a product rule in addition to the quotient rule, because of the product you have in your numerator $\left(x \tan x\right)$. But that's some way off. The first step for this problem is the quotient rule:

The formula for the quotient rule is:

$\frac{d}{\mathrm{dx}} \left[\frac{\textcolor{b l u e}{f \left(x\right)}}{\textcolor{red}{g \left(x\right)}}\right] = \frac{\textcolor{b l u e}{f ' \left(x\right)} \cdot \textcolor{red}{g \left(x\right)} - \textcolor{red}{g ' \left(x\right)} \cdot \textcolor{b l u e}{f \left(x\right)}}{\textcolor{red}{g} \left(x\right)} ^ 2$

So, we know that $f \left(x\right) = x \tan \left(x\right)$ and $g \left(x\right) = x + 4$. Now we just plug this into the equation:

$f ' \left(x\right) = \frac{\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(x \tan x\right)} \cdot \textcolor{red}{x + 4} - \textcolor{red}{\frac{d}{\mathrm{dx}} \left(x + 4\right)} \cdot \textcolor{b l u e}{x \tan x}}{\textcolor{red}{x + 4}} ^ 2$

And now, we just take a few derivatives.

$\frac{d}{\mathrm{dx}} \left(x + 4\right)$ is very straightforward -- the derivative of any linear function is it's slope, and $x + 4$ has a slope of 1, so:

$\frac{d}{\mathrm{dx}} \left(x + 4\right) = 1$

Now, $\frac{d}{\mathrm{dx}} \left(x \tan x\right)$ is a wee bit tricker because, as we mentioned before, you'll need to invoke the product rule to take this derivative.

This is as follows:

$\frac{d}{\mathrm{dx}} \left(\textcolor{b l u e}{f \left(x\right)} \ast \textcolor{g r e e n}{g \left(x\right)}\right) = \textcolor{b l u e}{f ' \left(x\right)} \textcolor{g r e e n}{g \left(x\right)} + \textcolor{g r e e n}{g ' \left(x\right)} \textcolor{b l u e}{f \left(x\right)}$

We know that $f \left(x\right) = x$ and $g \left(x\right) = \tan \left(x\right)$ (It doesn't matter which you pick to be which so long as you are consistent). So we just plug in:

$\implies \frac{d}{\mathrm{dx}} \left(\textcolor{b l u e}{\tan \left(x\right)} \ast \textcolor{g r e e n}{x}\right) = \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(\tan x\right)} \textcolor{g r e e n}{\left(x\right)} + \textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left(x\right)} \textcolor{b l u e}{\tan \left(x\right)}$

$= \left({\textcolor{b l u e}{\sec}}^{2} \left(x\right) \cdot \textcolor{g r e e n}{x}\right) + \left(\textcolor{g r e e n}{1} \cdot \textcolor{b l u e}{\tan} \left(x\right)\right)$

$= x {\sec}^{2} \left(x\right) + \tan \left(x\right)$

Now we plug this back into our quotient rule. This leaves us with:

$f ' \left(x\right) = \frac{\textcolor{b l u e}{\left[x {\sec}^{2} \left(x\right) + \tan \left(x\right)\right]} \cdot \left(\textcolor{red}{x + 4}\right) - \left[\textcolor{red}{\left(1\right) \cdot \textcolor{b l u e}{x \tan x}}\right]}{\textcolor{red}{x + 4}} ^ 2$

Some quick simplification:
$f ' \left(x\right) = \frac{\left(x + 4\right) \left[x {\sec}^{2} \left(x\right) + \tan \left(x\right)\right] - x \tan \left(x\right)}{x + 4} ^ 2$