Question

Can someone please help step by step?
d/dx [ ln(1- e^2x) / e^x^2]

Answer 1

Do you remember this one?\[\Large\bf \frac{d}{dx}\ln x\quad=\quad ?\]

Answer 2

I do not. But is that just x?

Answer 3

Oh boy.. you have a bunch of previous questions involving the natural log.
I would hope you have that one memorized by now :(

Ok so this is an important one to remember:\[\Large\bf \frac{d}{dx}\ln x\quad=\quad \frac{1}{x}\]That will help us when we need to take the derivative of the numerator.

Answer 4

\[\large\bf \left(\frac{\ln(1- e^{2x})}{e^{x^2}}\right)'\quad=\quad \frac{\color{#DD4747}{(\ln(1- e^{2x}))'}e^{x^2}-\ln(1-e^{2x})\color{#DD4747}{(e^{x^2})'}}{(e^{x^2})^2}\]

Answer 5

So we'll need to start with the `Quotient Rule`.
There is our setup.

Answer 6

ok. Im with you so far

Answer 7

So let's see if we can take the derivative of the first red part.

Answer 8

\[\Large \left(\ln[\color{#DD4747 }{stuff}]\right)'\quad=\quad \frac{1}{\color{#DD4747 }{stuff}}(\color{#DD4747 }{stuff})'\]So after we apply the rule for differentiating natural log, we'll have to apply the chain rule, multiplying by the dervative of the inner function.

Answer 9

\[\Large \left(\ln[\color{#008353 }{1-e^{2x}}]\right)'\quad=\quad \frac{1}{\color{#008353 }{1-e^{2x}}}(\color{#008353 }{1-e^{2x}})'\]

Answer 10

ohh ok.

Answer 11

I thought ln 1 was zero/

Answer 12

Mmm well we have a bunch of other stuff in the log with the 1, so we can't just pull apart the ln1 from it :\

Answer 13

ohh that makes sense.

Answer 14

with you now

Answer 15

So applying the chain rule, we need the derivative of that thing with the prime on it.\[\Large (1-e^{2x})'\quad=\quad (0-e^{2x}(2x)')\]

Answer 16

We have yet another chain to be applied D:
The exponential gives us the same thing back, but with the chain rule applied.

Answer 17

oh ok. Sorry for delay

Answer 18

What is the next step?