Expert: Paul Klarreich - 11/29/2007

Question
Hello,
I'm having trouble with a calculus homework question. The answer that I derived is not one of the available answer choices.

Find the derivative of f when
f(x) = 3 arctan (e^(2x)) + 2e^(-2x)

The answer that I got from this question is
6/(1+e^(4x)) * (e^(2x) - 2e^-2x)

Can you please help me with this?

Thanks,
Danielle

Answer
Questioner:   Danielle
Category:  Calculus
Private:  No
 
Subject:  Derivative of inverse trig functions and e^x
Question:  Hello,
I'm having trouble with a calculus homework question. The answer that I derived is not one of the available answer choices.

Find the derivative of f when
f(x) = 3 arctan (e^(2x)) + 2e^(-2x)

The answer that I got from this question is
6/(1+e^(4x)) * (e^(2x) - 2e^-2x)

Can you please help me with this?

Thanks,
Danielle
 
.............................

Hi, Danielle,

The trick is not to do too much at a time.  Some chain rule stuff you can do in youf head; some you should work out.

First term:

3 arctan (e^(2x)) = 3 arctan u,

where u = e^(2x), du/dx = 2e^(2x)
           3  
D(...) = ------- 2e^(2x)
        1 + u^2   

  6 e^(2x)
= ----------
 1 + e^(4x)



D(2e^(-2x)) = -4e^(-2x)

The whole thing is:

  6 e^(2x)
= ---------- - 4e^(-2x)
 1 + e^(4x)

That's not the same as your answer.  If you work at it a litle:


  6 e^(2x) - 4e^(-2x)(1 + e^4x)
= -----------------------------
   1 + e^(4x)


  6 e^(2x) - 4e^(-2x) - 4 e^2x
= -----------------------------
   1 + e^(4x)

  2e^(2x) - 4e^(-2x)
= --------------------
   1 + e^(4x)

How does that look?