Question

Differentiate the function. g(x)=ln (a-x/a+x) I'm confused, how do you start this problem?

Answer 1

\[\frac{d}{dx} (\ln (x)) = \frac{1}{x} \cdot \frac{d}{dx} (x)\]

Answer 2

In this case our \(x\) = \(\large \frac{a-x}{a+x}\) so..\[\frac{d}{dx} (ln(x)) = \frac{1}{\frac{a-x}{a+x}} = \frac{a+x}{a-x}\] the next part is taking the derivative of all that stuff.

Answer 3

You can either use the quotient rule: \(\frac{f'g -g'f}{g^2}\) or the product rule : \(f'g + g'f\) in solving these functions where \(f = a-x\) and \(g=a+x\)

Answer 4

Could you write g as ln(a-x)-ln(a+x) then differentiate?

Answer 5

Ooo there's another clever idea c:

Answer 6

possibly, yeah.

Answer 7

My initial thought was to go chain rule into quotient rule also lol.
Boy what a mess though :3

Answer 8

hmm I'll try the quotient rule and I'll see what I get :)

Answer 9

\[(a-x)(a+x)^{-1}\]\[f'g + g'f = -1(a+x)^{-1} -(a+x)^{-2}(1)(a-x) =-\frac{-1}{(a+x)} -\frac{(a-x)}{(x+x)^2}\] Theres a start.

Answer 10

Whoops, typo.

Answer 11

i the second fraction I meant too write \(\large -\frac{(a-x)}{(a+x)^2}\) Sorry.

Answer 12

\[\frac{-(a+x)-(a-x)}{(a+x)^2} =\frac{-a-x-a+x}{(a+x)^2}= \frac{-2a}{(a+x)^2}\]

Answer 13

So altogether you would have \[\frac{d}{dx} (\ln(x)) = \frac{a+x}{a-x} \cdot \left(-\frac{2a}{(a+x)^2}\right)\]

Answer 14

Quick question, in the product rule I got the first part -1(a+x)^-1 but how did you get -(a+x)^-2?

Answer 15

taking the derivative of the function using the formula \(nx^{n-1}\)

Answer 16

We brought \((a+x)\) up from where it was, in the denominator, therefore we have to raise it to the -1 power.

Answer 17

oh ok haha I forgot about that

Answer 18

Sort of like \(x^{-1} = \frac{1}{x}\) or \(\frac{1}{x} = x^{-1}\)

Answer 19

So if we have \(x^{-1}\), then taking it's derivative would be \(-1x^{(-1-1)}= -x^{-2}\)

Answer 20

oh ok I think that's what was making me get the wrong answer. Thanks for the help I really appreciate it :D

Answer 21

No problem :)

Answer 22

Once you're really good at using the chain rule of log functions, you should definitely try @myininaya 's way of finding the derivative using logarithmic properties :D

Answer 23

I'll keep that in mind thank you once again :)