Nezenic-differentiation of exponential functions

[Becky4paws]

you last said (May3rd) that I needed to use the chain rule as well as the product rule on this problem...
Here is my next attempt.

f(x) = (x+1)e^-2x for where x = 0
f(x) = (x+1)(e^-2x)(-2)
f(x) = -2(x+1)(e^-2x)
f(x) = (-2x-2)/e^2x

If that is correct, then I will move on to the product rule to find f'(x).

 

[stapel]

Note to tutors: The poster appears to be referring to this thread.

Note to poster: In the future, please post follow-ups to the original thread.

Also, please clarify what you are doing in this post. It appears that you are rearranging the function f(x) (since there appears to be no derivative), but the new version of the function is not equivalent to the old.

Please also specify which part of the complete worked solution provided in the other thread is insufficient to your needs.

Thank you.

Eliz.

 

[Steem]

f(x) = (x+1)e^-2x for where x = 0
f(x) = (x+1)(e^-2x)(-2)
f(x) = -2(x+1)(e^-2x)
f(x) = (-2x-2)/e^2x

f'(x)= -2(x+1)e^-2x + e^-2x

f'(x)= e^-2x(-2x-2+1) Don't forget that 1 is left after you factor out the e^-2x.

f'(x)= (-2x-1)/e^2x

A small algebraic mistake but good otherwise!

 

[Becky4paws]

Ok - I am still confused.
I thought I was supposed to use the chain rule to find a different way to write f(x), then I would use the product or quotient rule to find f'(x).

 

[galactus]

Hey Becky:

You use the chain rule when you differentiate something, not just to manipulate it algebraically.

You can use either the quotient rule or the product rule.

Product rule, leave it as \(\displaystyle \L\\(x+1)e^{-2x}\)

Quotient rule, write it as \(\displaystyle \L\\\frac{x+1}{e^{2x}}\)

Quotient rule: \(\displaystyle \L\\\frac{e^{2x}(1)-(x+1)(2e^{2x})}{e^{4x}}\)

Now, simplify. You should the same result using the product rule.