Question

Question #68962

Answer 1

Okay, this one is officially tedious (see explanation)

Explanation:

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

`d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)`

So for the first derivative, we'll let `f(x) = 4e^(2x)` and `g(x) = sec x`

`f'(x) = 8e^(2x)` (we used the chain rule here)

and `g'(x) = d/dx 1/cosx = sin^2x/cos^2x = sec x tan x`

put all this together, and you get:

`8e^(2x)secx + 4e^(2x)secx tanx`

and to be fastidious, you can factor out term `4e^(2x)secx`,
making your first derivative:

`4e^(2x)secx(2 + tanx)`

Now the tedious part - the second derivative. Once more, use the rule for the product of 2 functions. This time, we let `f(x) = 4e^(2x)secx` and `g(x) = (2 + tanx)`

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

`g'(x) = d/dx (2 + tanx) = sec^2 x`

Put all these together, and we have:

`(d^2x)/dx^2 4e^(2x)secx =`

`(8e^(2x)secx + 4e^(2x)secx tanx)(2 + tanx) + 4e^(2x)secx(sec^2x)`

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

`16e^(2x)sec x + 8e^(2x)sec x tan x + 8e^(2x)sec x tan x + 4e^(2x)sec x tan^2 x + 4e^2x sec^3 x`

`= 16e^(2x)sec x + 16e^(2x)sec x tan x + 4e^(2x)(sec x tan^2 x + sec^3 x)`

...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