Question

find the derivative f(x)=log3x

Answer 1

\[f(x)=\log _{3}x\]

Answer 2

I'm beginning to think that you're a math robot

Answer 3

bleep blorp...i mean...crap I've been discovered lol

Answer 4

Seriously!

Answer 5

Use the rule that

If

\[\Large f(x) = \log_{b}(x)\]

then

\[\Large f^{\prime}(x) = \frac{1}{x*\ln(b)}\]

Answer 6

So in this case, b = 3

Answer 7

again this is just another rule to learn...but it's very similar to the rule that

If \[\Large f(x) = \ln(x)\] then \[\Large f^{\prime}(x) = \frac{1}{x}\]

Answer 8

okay so 1/xln3

Answer 9

so that's something to connect

Answer 10

so x*1/3?

Answer 11

x/3

Answer 12

yes, in this case, \[\Large f^{\prime}(x) = \frac{1}{x*\ln(3)}\]

Answer 13

no you can't get rid of the ln that easily

Answer 14

dang!

Answer 15

yeah I wish you could, but you can't

Answer 16

maybe it's prudent to show where that came from?
After all...
\[\huge \log_3x=\frac{\ln \ x}{\ln \ 3}\]
And now, the
\[\huge \frac1{\ln 3}\]is just another constant.

Answer 17

good point terenzreignz, that shows how the two connect

Answer 18

alright, so this boils down to the derivative being 1/xln3

Answer 19

I cant simplify or anything?

Answer 20

yes \[\Large f^{\prime}(x) = \frac{1}{x*\ln(3)}\]

Answer 21

^ That's as simple as it can be :)

Answer 22

first use change of base then use natural log
log3x can be written as using change of base property of log
\[\Large \log_{3}(x)=\frac{\ln(x)}{\ln(3)}\]
now take the constant ln(3) out
\[\Large \frac{d}{dx}(\frac{\ln(x)}{\ln(3)})=\frac{1}{\ln(3)}*\frac{d}{dx}(\ln(x)\]
\[\Large \frac{1}{\ln(3)}*\frac{1}{x}=\frac{1}{\ln(3)x}\]

Answer 23

you can rearrange terms I guess, but it's not going to get any simpler

Answer 24

alright...that seems okay

Answer 25

not too hard, until i look at the next set and theres RADICALS!

Answer 26

Radicals are just fractional exponents... just sayin' ;)

Answer 27

\[f(x)=2^{x}\sqrt{x}\]

Answer 28

so that changes to 1/x?

Answer 29

Use the product rule, but remember that
\[\huge f(x)=2^x \cdot x^{\frac12}\]

Answer 30

how does that apply to the fraction though

Answer 31

the radical i mean

Answer 32

Power rule :)
\[\huge \frac{d}{dx}x^m=mx^{m-1}\]for any real number m.

Answer 33

So I can use that for the 2^x and make it 2x but I still have that times the radical x

Answer 34

No!
The power rule applies when x is the BASE, not when x is the EXPONENT. We'll get to that later. For now, what's the derivative of the square root of x?
\[\huge \frac{d}{dx}\sqrt x=\frac{d}{dx}x^{\frac12}=?\]

Answer 35

1/x?

Answer 36

Nope...Remember the power rule...\[\huge \frac{d}{dx}x^m=mx^{m-1}\]and take m = 1/2

Answer 37

1/2(radical x)^-1/2

Answer 38

This?
\[\huge \frac12 \sqrt x^{-\frac12}\]?

Answer 39

yes

Answer 40

Well, in that case, no. We already converted the radical to a fractional exponent, why did you put it back? :P

Answer 41

I guess I missed that part!

Answer 42

So... care to modify your answer? :P

Answer 43

You were *almost* right, though.

Answer 44

so just take out the radical

Answer 45

Yeah :D
\[\huge \frac12x^{-\frac12}=\frac1{2\sqrt x}\]
Ok? :P

Answer 46

okay!

Answer 47

So, no questions on the part of...
\[\huge \frac{d}{dx}\sqrt x = \frac{1}{2 \sqrt{x}} \]

Answer 48

well now that i am able to see where you got the 1/2 from I understand

Answer 49

So, shall we proceed? :)

Answer 50

absolutely

Answer 51

Now, work on the derivative of
\[\huge 2^x\]Any idea how to proceed?