Question

Differentiate the following function with respect to x

(1) In2x

Answer 1

1/x

Answer 2

\[\frac{d}{dx}\ln(ax)=\frac{1}{x}\] for any constant a

Answer 3

I'm not sure of the rules of In... Are they the same or different than e?

Answer 4

that is because
\[\ln(ax)=\ln(a)+\ln(x)\]
the derivative of
\[\ln(a)\] is zero, because it is a constant, and the derivative of
\[\ln(x)\] is
\[\frac{1}{x}\]

Answer 5

\[ln(x)=\log_e(x)\]

Answer 6

Hm Ok.. so, what would In(2x-1) be?

Answer 7

\[\frac{1}{2x-1}\times \frac{d}{dx}[2x-1]\] by the chain rule, or
\[\frac{2}{2x-1}\]

Answer 8

2/(2x-1)

Answer 9

in general the derivative of
\[\ln(g(x))\] is
\[\frac{g'(x)}{g(x)}\]

Answer 10

you will use this repeatedly, so it is good to get it under your belt now

Answer 11

\[what~ does~ ~g'(x) ~mean?\]

Answer 12

(lnu)' = u'/u
(ln 2x)' = 2/2x = 1/x
In(2x-1)' = 2/(2x -1)

Answer 13

note that this also says that since
\[\frac{d}{dx}\ln(g(x))=\frac{g'(x)}{g(x)}\]this also means we can solve for g' to get
\[g(x)\times \frac{d}{dx}\ln(g(x))=g'(x)\] a good way to find the derivative sometimes

Answer 14

g(x) = u

Answer 15

\[g'\] means the derivative of g

Answer 16

Hmm. Ok

Answer 17

I've done some more on my own, and have managed, but what's\[ In {1 \over x}?\]

Answer 18

-1/x

Answer 19

Yes, but why?

Answer 20

\[y=\ln\frac{1}{x}\]
\[\frac{dy}{dx}=-\frac{1}{(\frac{1}{x})}(x ^{-2})\]
\[\frac{dy}{dx}=-x(x ^{-2})\]
\[\frac{dy}{dx}=-\frac{x}{x ^{2}}\]
\[\frac{dy}{dx}=-\frac{1}{x}\]

Answer 21

Oh, That makes soo much more sense! Thanks so much!