What's Wrong With This Answer?
1 sec, posting the attached img.
What's Wrong With This Answer?
1 sec, posting the attached img.
@Mathematics

Question

What's Wrong With This Answer?
1 sec, posting the attached img. 
 What's Wrong With This Answer?
1 sec, posting the attached img. 
 @Mathematics

anonymous · Answer

attachment

anonymous · Answer

I double checked this, and I still don't see where the heck I went wrong.

anonymous · Answer

did you take the derivaitve of the e^-x twice

anonymous · Answer

i think you did

anonymous · Answer

Where?

anonymous · Answer

when you applied the product rule

anonymous · Answer

No because that second -ve sign comes from the cos

anonymous · Answer

because (cosx)' = -sinx

anonymous · Answer

looks good to me if this is on line work, you probably have to write 
sin(pix) rather than sinpix

anonymous · Answer

Aw these stupid technical errors.

anonymous · Answer

anonymous · Answer

Why is the derivative log(2) in that link instead of ln(2)?

zarkon · Answer

read the fine print...log is the natural log

anonymous · Answer

the large print giveth and the small print taketh away

anonymous · Answer

that wolfram link is not the question you were asked

anonymous · Answer

If you mean the wolfram solution, it is ln(2)
"log (x) is the natural logarithm"