I was wondering if someone could make me a list of how to find the derivatives of stuff with "e" and "LN"

I know how to add, subtract, power rule, product rule, chain rule etc.

I am not sure if I know exactly what I am asking, but I am having trouble finding "harder" derivatives. Not the regular (x+5)^5 or something. I can do the stuff with a regular "x" but when ln, log, and e and stuff like that gets mixed in, I get so confused.

I think what I am asking is finding stuff that looks like this. ( I am not sure what these are called, so I am not sure what I should be googling to look for help)

f(x) = ln[(1 + x)(1 + x2)2(1 + x3)3 ]

f (x) = 2^x

f(x) = x log(2x)

f(x)=e^2x + 3

Because I can't find the derivatives of the "harder" stuff, I am having trouble finding the Antiderivatives and integrating stuff.


Which leads me to have trouble doing things like f(x,y) and then taking fxx, fyy, and fxy .

Can someone please guide me to where I should begin looking for extra help besides the textbook? I have the "rules" in the textbook such as "The derivative of ex with respect to x is equal to e^x." but I have NO IDEA how that applies to actual problems and numbers. I am having a hard time seeing the correlation between the defn and actual problems.

It's been a gazillion years since I've looked at this stuff (OK, maybe like 10-12), but some of your problems look like chain rule examples. I got a nice wikipedia article on "chain rule," though it's a bit dense to read.

While I can't solve these off-hand right now, I'm sure it wouldn't take me long to brush up on calc and figure them out (once I teach myself stats this afternoon!) I think I'm *slightly* better at math than my fiance, but I'll run this by him in case he can solve them off-hand. I can get back to you in a couple of days if you'd like, but meanwhile I think "chain rule" will help. Good luck!

Edit-- sorry, I just noticed you said you're ok with chain rule. In that case, treat "e" and "ln" the same way you'd treat any other function in chain rule. Since derivative of e^x = e^x, then if you have e^(2x) the two functions composing it are:

f=e^(g(x))
g= 2x

Then you have derivatives as follows: f'= e^2x and g'=2, so by chain rule the derivative of the whole thing is 2*e^(2x). Does that make sense? (And I *really* hope I got that right doing it on the fly after so long!).

Again, I can get back to you on some of the other examples in a while.

I think I've figured these out (can you confirm that the 1st example is correct? It looks like there's an operator missing before the central "2").

It's late here so I can't do them out for you now, but I can come back to it tomorrow. It'll be easier for me to help you if you can post how far you've gotten with these, or what your thoughts are on how to go about solving them.

I think I am having troubles solving derivatives for things like e^what ever power, LNx+2 etc....

There has to be a list of "rules" somewhere right?
I tried looking in my textbook, and it isn't much help.

I am not sure if the one you did is correct...haili some help?

I'll try to help, I just finished my calculus final and this stuff is pretty fresh in my head.
Sorry if this is not what you're looking for, it wasn't too clear what you were confused about, but I'll try the best I can

Derivative of ln
example: f(x)=ln(x+2)
first you take the derivative of ln, which is 1/(x+2)
then you MUST take the x+2 and find the derivative of that (which is 1) and multiply it to your 1/(x+2)
So f'(x)=1/(x+2)

f(x)=ln(3x)
again, find the derivative of LN first, which is 1/3x
THEN, the inner function which is 3x, find the derivative of that (3). You must multiply that with your 1/3x
So f'(x)=3/3x which simplifies to 1/x

y=2^x
take the ln of both sides!
Remember in ln, the power goes down to the front? Well, it's happening now.
ln y = x ln 2
1/y * dy/dx = ln 2
dy/dx = y ln 2
Then, you just pop the y back in, which is 2^x.
dy/dx = 2^x ln 2

Derivative of E
remember, derivative of e^x is e^x!!
but when you have something else attached to it, such as e^(5x), take the derivative of 5x, which is 5. Multiply that onto your e^(5x). That gives you f'(x)=5e^(5x)

I hope I didn't confuse you even more... I tried to make it as simple as I can since it's hard to explain math problems over the internet. Good luck on your final!!!

Here are some natural logarithm rules:
ln(xy) = ln(x) + ln(y)
ln(x/y) = ln(x) - ln(y)
ln(x^r) = r ln (x) for any real number r. I found this one to be really useful.
Some other things to remember:
ln(x)= y is equivalent to saying x = e^y for real numbers x>0 and y
ln(e^x) = x for all real numbers
e^ln(x) = x for every x >0

Derivative of logb (x) = 1/ xln(b)
Derivative of ln (x) = 1/x
Derivative of e^x = e^x
Derivative of a^x = a^x ln (a)

I worked out your first problem with as much detail as I could. I can do more tomorrow with my tablet, if you want to see all the steps to the problems.
tinypic.com . . .

This might be unpopular advice, but when I tutored math students, I always encouraged them to memorize as few "rules" as possible and work things out from first principles. The people I know who are truly good at math have memorized very few rules (and I'm talking about people who have done all of calculus, multivariable calculus, and linear algebra/ differential equations and have successfully gone on to do even more).

In these examples, you really need to know only that derivative of lnx= 1/x, and derivative of e^x= e^x.

Now, when you have e^f(x), I usually first just consider it as "e^(stuff)." Then there's only 2 steps:

1) derivative of e^(stuff) = e^(stuff), so write that as the 1st part of the answer
2) now multiply it by the derivative of the "stuff"

And you're done!

Same process applies for ln's, trig functions etc.

Edit-- I checked that the one I did above is correct. If you can figure out for yourself why you thought it might be incorrect, that might give you a lead on what you need to brush up on.