These are the last three problems that I need to know before the test .
Find f(x)'
A) f(x) = log (1-x)^2
B) f(x) = log9(2^x + 1)
C) f(x) = 2^(log(3)x)
The 9 is subscript in problem B
The 3 is subscript in problem C
Thanks alot for the help.
You definitely need to know how to find the derivative of $\displaystyle \log_{a}x$. Using a change of base this becomes $\displaystyle \frac{\ln(x)}{\ln(a)}$ and now this can be differentiated to get $\displaystyle \frac{1}{x\ln(a)}$.
For the first two, they aren't that bad just use the chain rule. The third one you can use that if $\displaystyle f=a^x$, then $\displaystyle f'=a^x \ln(a)$
Edit: whoops, sorry Jameson, didn't see that you had posted again
I haven't seen this notation before, and I am assuming that f(x)' is the derivative of f(x). If this is not the case, please ignore me.These are the last three problems that I need to know before the test .
Find f(x)'
Remember the formula $\displaystyle \frac{d}{dx}\log(g(x)) = \frac{g'(x)}{g(x)}$ and/or use the chain rule.A) f(x) = log (1-x)^2
Use $\displaystyle \log_a(b) = \frac{\log(b)}{\log(a)}$ and $\displaystyle a^x = e^{\log(a)x}$B) f(x) = log9(2^x + 1)
Use all the formulas from the other 2 questions.C) f(x) = 2^(log(3)x)
a)$\displaystyle f=(\log (1-x))^2$
Now like I told you, convert the log base 10 to log base "e".
$\displaystyle f= \left( \frac{\ln(1-x)}{\ln(10)} \right) ^2$
$\displaystyle f' = 2 \left( \frac{\ln(1-x)}{\ln(10)} \right) * \left( \frac{-1}{(1-x)\ln(10)} \right)$
Now I fear after all this that your problem was actually $\displaystyle f=\log[(1-x)^2]$. Which one is it?
no..? isnt the rule of logs : f(x)=log a (f(x)) , where a is a subscript. therefore f ' (x)= 1 1
___ . ___ . f ' (x)
ln a f(x)
sorry if you dont understand what i wrote..
so basically, for what you do is :
f ' (x) = 1 over (ln 10) times 1 over (1-x)^2 times 2(1-x)times(-1)
hope this helps..