differentiate y=log2 x
\[y= \log_{2} x\]
Hey Dan, remember your change of base formula? :)\[\large\rm \log_b(a)=\frac{\ln(a)}{\ln(b)}\]
yes but i dont think i understand how to do it. I get lnx/ln2 and then i dnot know what to do after that
Ok great. Let's pull the numerical portion out front,\[\large\rm \frac{\ln x}{\ln2}\quad=\quad \left(\frac{1}{\ln2}\right)\ln x\]
Even though this coefficient looks very strange and complicated, it's just a NUMBER. Think of it like this,\[\large\rm (5)x\]You remember how to differentiate that, yes?
how did you pull the numerical portion out i dont understand that
\[\large\rm \frac{1}{\ln2}\cdot \ln x\quad=\quad \frac{1}{\ln2}\cdot\frac{\ln x}{1}\quad=\quad \frac{1\cdot \ln x}{\ln2\cdot 1}\]If you follow these steps in reverse, then it might make some sense. I split it into two fractions, using fraction multiplication. You have to artificially create a 1 in the top and bottom for it to work out nicely.
Woops I meant to say (5)ln x in the previous comment, no big deal though.
okay
\[\large\rm \frac{d}{dx}(\text{#})\ln x\]So we have some weird looking number times our log. Remember how to differentiate your natural log? :)
yes i think
i;m not sure how to continue
What is the derivative of ln x?
1/x
1/x (1)
Ok good. So if we had 5lnx, we we would end up with 5(1/x) right?
yes
The 5 can come outside of the differentiation process since it is a constant coefficient.\[\large\rm \frac{d}{dx}5\ln x=5\frac{d}{dx}\ln x=5\frac{1}{x}\] Same thing happens with our problem,\[\large\rm \frac{d}{dx}\left(\frac{1}{\ln2}\right)\ln x=\left(\frac{1}{\ln2}\right)\frac{d}{dx}\ln x=?\]
(1/ln2)(1/x)
Yayyy good job! We can combine the fractions as a last step perhaps, just so it looks a little nicer,\[\large\rm =\frac{1}{x \ln2}\]
thank youuu, i will work on practice problems for this to see if i understand it better I will be back if i am stuck
cool
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