# Question #68962

Oct 31, 2017

Okay, this one is officially tedious (see explanation)

#### Explanation:

We can use the rule for finding the derivative of the product of 2 functions:

$\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So for the first derivative, we'll let $f \left(x\right) = 4 {e}^{2 x}$ and $g \left(x\right) = \sec x$

$f ' \left(x\right) = 8 {e}^{2 x}$ (we used the chain rule here)

and $g ' \left(x\right) = \frac{d}{\mathrm{dx}} \frac{1}{\cos} x = {\sin}^{2} \frac{x}{\cos} ^ 2 x = \sec x \tan x$

put all this together, and you get:

$8 {e}^{2 x} \sec x + 4 {e}^{2 x} \sec x \tan x$

and to be fastidious, you can factor out term $4 {e}^{2 x} \sec x$,

$4 {e}^{2 x} \sec x \left(2 + \tan x\right)$

Now the tedious part - the second derivative. Once more, use the rule for the product of 2 functions. This time, we let $f \left(x\right) = 4 {e}^{2 x} \sec x$ and $g \left(x\right) = \left(2 + \tan x\right)$

Note that we just calculated $f ' \left(x\right)$ - it's the first derivative of the original function.

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

Put all these together, and we have:

$\frac{{d}^{2} x}{\mathrm{dx}} ^ 2 4 {e}^{2 x} \sec x =$

$\left(8 {e}^{2 x} \sec x + 4 {e}^{2 x} \sec x \tan x\right) \left(2 + \tan x\right) + 4 {e}^{2 x} \sec x \left({\sec}^{2} x\right)$

...which is your answer, but we need to multiply out, collect terms, and simplify to the extent that we can:

$16 {e}^{2 x} \sec x + 8 {e}^{2 x} \sec x \tan x + 8 {e}^{2 x} \sec x \tan x + 4 {e}^{2 x} \sec x {\tan}^{2} x + 4 {e}^{2} x {\sec}^{3} x$

$= 16 {e}^{2 x} \sec x + 16 {e}^{2 x} \sec x \tan x + 4 {e}^{2 x} \left(\sec x {\tan}^{2} x + {\sec}^{3} x\right)$

...since we live in 2017, we can double check this with Wolfram Alpha (a task I recommend to you). It looks like it agrees, so we're done. If you need me, I'll be in the lounge.

GOOD LUCK