# How do I find the derivative of the function y = sin(tan(4x))?

Jan 26, 2015

You have a composite function, which are to be derived using the chain formula. If you're in the case $f \left(g \left(x\right)\right)$, the chain formula says that $f \left(g \left(x\right)\right) ' = f ' \left(g \left(x\right)\right) \setminus \cdot g ' \left(x\right)$. Your case is slightly different, since you have three functions: you're in the case $f \left(g \left(h \left(x\right)\right)\right)$, where $f \left(x\right) = \setminus \sin \left(x\right)$, $g \left(x\right) = \setminus \tan \left(x\right)$, and $h \left(x\right) = 4 x$.

In these cases, you simply need to derive the most "external" function, and then multply the result by the derivative of the internal one, and so on.

First of all, let's remark the derivatives of your three functions: you have that
$\frac{d}{\mathrm{dx}} \setminus \sin \left(x\right) = \setminus \cos \left(x\right)$,
$\frac{d}{\mathrm{dx}} \setminus \tan \left(x\right) = \frac{1}{\setminus {\cos}^{2} \left(x\right)}$, and
$\frac{d}{\mathrm{dx}} 4 x = 4$.

Let's calculate f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x): we have
$\setminus \cos \left(\setminus \tan \left(4 x\right)\right) \setminus \cdot \frac{1}{\setminus {\cos}^{2} \left(4 x\right)} \setminus \cdot 4$, which we can rewrite as
$4 \frac{\setminus \cos \left(\setminus \tan \left(4 x\right)\right)}{\setminus {\cos}^{2} \left(4 x\right)}$ or again, if you prefer, $4 \setminus \cos \left(\setminus \tan \left(4 x\right)\right) \setminus \cdot \setminus {\sec}^{2} \left(4 x\right)$

Jan 26, 2015

$\frac{d}{\mathrm{dx}} \sin \left(\tan \left(4 x\right)\right) = 4 \cos \left(\tan \left(4 x\right)\right) {\sec}^{2} \left(4 x\right)$

Explanation

You would need to use the chain rule to find this derivative, because you are dealing with composite functions. Composite functions are functions where one (or more) functions are composed in another function. You can think of it as having an "outer" function with an "inner" function composed in it.

Here, finding the inner and outer functions is quite straight forward. In $y = \sin \left(\tan \left(4 x\right)\right)$ You can see than the $\tan \left(4 x\right)$ is inside the $\sin$- so the $\sin$ is a function of $\tan \left(4 x\right)$. But you have to be careful with this problem because there is another composite function. The $4 x$ is yet another composite function, composed inside the $\tan$. Since there are two composite functions, you have to do the chain rule twice.

In words, the chain rule says:

The derivative of the outer function (with the inner function left alone) times the derivative of the inner function

In this case, we expand it to derive the second composite function (after deriving its outer function left alone).

$\frac{d}{\mathrm{dx}} \sin \left(\tan \left(4 x\right)\right)$

Step 1)
First we need to derive the $\sin$ alone, ignoring the $\tan \left(4 x\right)$ on the inside;
$\frac{d}{\mathrm{dx}} \sin \left(\tan \left(4 x\right)\right) = \cos \left(\tan \left(4 x\right)\right)$

NOTE 1: This isn't the completed derivation problem, I'm just showing a combination of steps that I will combine in the end- So I'm leaving deriving the inner functions in seperate steps.

Step 2)
Okay, now we need to derive the first inner function, that is $\tan \left(4 x\right)$ But again, leave its inner function alone for now.

$\frac{d}{\mathrm{dx}} \tan \left(4 x\right) = {\sec}^{2} \left(4 x\right)$

Step 3)
Finally, we derive the last inner function.

$\frac{d}{\mathrm{dx}} 4 x = 4$

Step 4)
NOTE 2: This is where we do the actual correct derivation, by combining the steps

d/dx sin(tan(4x))= step 1 times step 2 times step 3= $4 \cos \left(\tan \left(4 x\right)\right) {\sec}^{2} \left(4 x\right)$