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Mathematics 21 Online
OpenStudy (anonymous):

Can you help me understand how to differentiate involving logarithms? Please see the Q for an example...

OpenStudy (anonymous):

\[f(x) = x^2\log_{2} (x^2+1)\]

OpenStudy (amistre64):

the log2 is best viewed as ln(x^2+1)/ln(2)

OpenStudy (amistre64):

since 1/ln(2) is a constant, all thats left is to consider this as a usual ln(u) type issue

OpenStudy (anonymous):

I understand to start by converting the log base 2 into a natural ln base e, which leads to: \[f(x)=x^2(\ln (x^2 + 1) /\ln 2)\] I am unsure how to address the fractional term involving the ln's.

OpenStudy (amistre64):

1/ln(2) is a constant; nothing changes about it ... so

OpenStudy (amistre64):

remove it, stick it to the side, put a stick note on top ot it and work the rest of the problem

OpenStudy (anonymous):

Ah, the 1/ln2 is a constant. Good insight. Thank you, @amistre64.

OpenStudy (anonymous):

what is a stick note?

OpenStudy (amistre64):

sticky note is what thats spose to be, a post it note, something to remind you not to forget about it in the end :)

OpenStudy (anonymous):

okay, got it.

OpenStudy (anonymous):

I think the key to this one for me was recognizing that constant.

OpenStudy (amistre64):

yep

OpenStudy (anonymous):

Would I then use the product rule against the remaining terms x^2 and ln(x^2+1) ?

OpenStudy (amistre64):

i would, seeing how its a product ;)

OpenStudy (anonymous):

Sometimes all the new stuff to learn can create uncertainty of the obvious.

OpenStudy (amistre64):

yeah, im dong that with existence and uniquenes thrms right now wondering what gppd to do and stuff

OpenStudy (amistre64):

good to do and stuff ... cant type

OpenStudy (anonymous):

What class are you taking at this time?

OpenStudy (amistre64):

im in college going to take linear algebra, calc3, and diffy qs in the spring

OpenStudy (anonymous):

I am taking Calc 1.

OpenStudy (anonymous):

And reviewing precalc daily!

OpenStudy (anonymous):

Thanks, best wishes.

OpenStudy (amistre64):

good luck :)

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