Differentiating natural logs and exponential functions?

Mirth

Hey all, I'm really having a hard time figuring out a couple of problems in which I have to differentiate:

1: [tex] \frac{e^3^x}{\ln x} [/tex]

I just don't know how to put it together... I know that [tex]e^3^x[/tex] is [tex]3e^3^x[/tex], and I know that you can't different [tex]\ln x[/tex], so I dunno what to do from there...

And:

2: [tex] \ln(e^-^2^x + e^-^x)[/tex]

Totally not sure what to do there... Any help would be appreciated, I've looked online for similar examples and couldn't really find anything relevant to help me. Thanks!

Hootenanny

Let's take the first question first,
Correct.
Erm... yes you can!

HINT: Let y = lnx, then x = e^y. Now differentiate x = e^y with respect to x.

Mirth

Hrm, I'm feeling a bit retarded so bear with me, hehe. So, [tex]\frac {1}{x}[/tex]?

Hootenanny

Correct

So as far as putting it all together, can you suggest anything? Perhaps some sort of rule...?

Mirth

I'm really retarded at math, ugh... Hehe. So I have [tex]\frac {3e^3^x}{1/x}[/tex] ... Pretty sure I'm lost on what to do from there. :(

Hootenanny

No that is not correct.

[tex]\frac{d}{dx} \frac{f\left(x\right)}{g\left(x\right)} \neq \frac{f^\prime\left(x\right)}{g^\prime\left(x\right)}[/tex]

What 'rules' of differentiation do you know?

Mirth

Not sure of any rules by name... I'm going to have to research more examples, I'm so rusty in math that I am unsure...

Hootenanny

Try searching for the quotient rule, or the product rule and chain rule.

Mirth

I kinda give up on the first one, heh... Can someone tell me if I have #2 right?:

Answer: [tex]\frac {1 - e^-^2^x}{e^-^x + e^-^2^x} [/tex]

Pengwuino

The quotient rule and product rule explains how to differentiate two functions that divide are divided by or multiplied by eachother. Just look that up and identify f(x) and g(x) (that is the standard notation at least).

For #2, it isn't correct. What did you do to come up with that answer?

slider142

Mirth, can you show us step by step how you differentiated that expression?

tiny-tim

Hi Mirth!
ok, do you recognise this one by face …

(fg)' = … ?