Calculus Help (Differentiation?)

[jennmaths]

If g(x)=e^(2x), (g(h+1)-g(h-1))/h equals?
The following image present my work so far. I got e^(4) but the answer should be 2e^(2).

I’d appreciate any help!

 

[Harry_the_cat]

2 things.
1. Is it g(1-h) or g(h-1)
2. 2(1+h) = 2+2h not 2+ h
Fix those things and try again

 

[jonah2.0]

Beer soaked comment follows.

If g(x)=e^(2x), (g(h+1)-g(h-1))/h equals?
The following image present my work so far. I got e^(4) but the answer should be 2e^(2).
View attachment 25677
I’d appreciate any help!

I like your handwriting a lot.
One of the best handwriting I've ever seen.

 

[Deleted member 4993]

Beer soaked comment follows.

I like your handwriting a lot.
One of the best handwriting I've ever seen.

Not only handwriting - presentation is excellent!!

 

[HallsofIvy]

If \(\displaystyle g(x)= e^{2x}\) then \(\displaystyle \frac{g(1+ h)- g(1- h)}{h}=\)
\(\displaystyle \frac{e^{2(1+h)}- e^{2(1- h)}}{h}= \frac{e^{2+ 2h}- e^{2- 2h}}{h}\)
\(\displaystyle = e^2\left(\frac{e^{2h}- e^{-2h}}{h}\right)\).

That is as simple as you can get it.

 

[jennmaths]

2 things.
1. Is it g(1-h) or g(h-1)
2. 2(1+h) = 2+2h not 2+ h
Fix those things and try again

1. It is g(1-h)
2. I had caught that and changed it the next step but forgot to rewrite it in the step before. It still leads me to e^(4).

 

[Dr.Peterson]

You appear to be thinking that if you take the log of a difference, the result is the difference of the logs, though you correctly take the log of a product or quotient (on the inside, but not the division by h).

In addition, your presentation is poor, as you keep putting equal signs between things that are not equal.

 

[lex]

Are you sure the question did not have 2h on the bottom?
Is the question to find the derivative from first principles, at x=1?

If [MATH]g(x) = e^{2x}[/MATH] then find [MATH]\lim\limits_{h \to 0}\frac{g(1+ h)- g(1- h)}{2h}[/MATH]