I need walked through this question: take the derivative of y=log2 3x

Question

amistre64 · Answer

base 2 each side

amistre64 · Answer

2^y = 3x

now solve as usual
$$[B^x]'=B^x~ln(B)$$

amistre64 · Answer

2^y ln(2) y' = 3

y' = 3/(2^y ln(2))

and since y = L.2(3x)

y' = 3/(2^(L.2(3x)) ln(2))

y' = 3/(3x ln(2))

y' = 1/(x ln(2))

laddiusmaximus · Answer

wait multiply two to both sides?

amistre64 · Answer

no, BASE 2 each side

amistre64 · Answer

you have a log, you can modify it to an exponent if need be

amistre64 · Answer

otherwise change of base the log and run the ln(3x) to fruition

amistre64 · Answer

$$L_b(a)=\frac{ln(a)}{ln(b)}$$$$L_2(3x)=\frac{ln(3x)}{ln(2)}$$

laddiusmaximus · Answer

the answer is log2e/x

laddiusmaximus · Answer

how do we get there?

amistre64 · Answer

thats not "the" answer.  that is just the answer they chose to "simplify" to.

amistre64 · Answer

http://www.wolframalpha.com/input/?i=d%2Fdx+log_2%283x%29 how they got to your answer, is anyones guess

laddiusmaximus · Answer

gotcha, thanks.

amistre64 · Answer

http://www.wolframalpha.com/input/?i=1%2F%28xln%282%29%29-log_2%28e%29%2Fx just so you know, the 2 are equal :)

anonymous · Answer

To get to the answer in the book:
$$y' = \frac{1}{x \ln 2} = \frac{\ln e}{x \ln{2}}$$

Since $\log_2 e = {\ln e \over \ln 2}$
$$\frac{1}{x}\frac{\ln e}{\ln 2} = \frac{\log_2 e}{x}$$

laddiusmaximus · Answer

not really seeing it at the moment, but ill do a few more to try and understand it. that, or just move on to intergrals.