stapel_eliz wrote:doodoo21 wrote:y= 2x + 3^(-12x)

y= x^(-2x) - (e^2x)/5

I need to find both the first and second derivative.

Where are you getting stuck?

Please be complete. Thank you!

From my understanding the first one has to do with ln

so dy/dx = 2 - 12ln3(3^-12x)

I'm not sure whether I multiply the -12ln3 with the (3^-12x). Even if that is correct or not, I don't understand how to get the second derivative as I'm not sure on how to derive logs >.<

The second one would incorporate the chain rule?

Therefore for x^-2x:

let y= x^u , u = -2x

dy/dx = u(x) . -2 = 4x^2

And for - (e^2x)/5:

let y= - (e^u)/5 , u= 2x

dy/dx = (e^u)/5 . 2 = (2e^2x)/5

I'm not sure on how to handle the /5 and on how to derive the first derivative, especially when they deal with exponential and logs.

If you could further explain where I've gone wrong, I would be grateful.