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Mathematics 16 Online
OpenStudy (anonymous):

Differentiate f(x) = 30 + 10e^-0.05x

OpenStudy (mimi_x3):

\[ f(x) = 30 + 10e^{-0.05x}\] where are you having problems?

OpenStudy (anonymous):

please help this confuses me i know the answer i just want to know how the answer is found.....

OpenStudy (mimi_x3):

\[ f'(x) = -0.05*10e^{-0.05}\] i dont know how to expplain though

OpenStudy (anonymous):

this question is in my tuttorial question

OpenStudy (mimi_x3):

\[f'(x) = -0.05*10e^{-0.05x}\]

OpenStudy (unklerhaukus):

\[f(x)=ax^n\] differentiating , be reduce the power of x in each term by one, multiplying the term by the old power \[f'(x)=anx^{n-1}\]

OpenStudy (unklerhaukus):

be*we

OpenStudy (unklerhaukus):

\[f(x)=g(x)+h(x)\] \[f'(x)=g'(x)+h'(x)\]

OpenStudy (unklerhaukus):

note that the power of x in a constant term is zero

OpenStudy (anonymous):

in the the answer here is \[f'(x)=-0.5*e^-0.05x\]

OpenStudy (mimi_x3):

yeah i got that as the answer.. or simply \[\frac{d}{dx} \left(e^{ax+b}\right) = ae^{ax+b}\]

OpenStudy (neo92):

when answering this u have to keep these in mind 1. differential of a constant is zero d(k)/dx = 0 , where k is a constant 2.\[de^{x}/dx = e^{x}\] genarally \[de^{ax^{n}}/dx = e^{ax^{n}}dax^{n}/dx\]

OpenStudy (anonymous):

thank you guys i see now ... to obtain -0.5 the 10 was multipled by -0.05

OpenStudy (anonymous):

this chain rule is not for me hey here is another sum\[f(x)=(x^2+3x+5)e^6x\]

OpenStudy (unklerhaukus):

\[f(x)=(x^2+3x+5)e^{6x}\] \[f'(x)=(x^2+3x+5)'e^{6x}+(x^2+3x+5)(e^{6x})'\]

OpenStudy (anonymous):

thank you very much

OpenStudy (unklerhaukus):

whats your final answer?

OpenStudy (anonymous):

\[f'(x)=(6x^2+20x+33)e^6x\]

OpenStudy (unklerhaukus):

man thats hard to do in your head, yeah correct

OpenStudy (unklerhaukus):

well done

OpenStudy (anonymous):

that yes....but i was never good in math

OpenStudy (unklerhaukus):

well now you are,

OpenStudy (anonymous):

how does one know that the chain rule is required to be used to solve for the answer

OpenStudy (anonymous):

lol guess so

OpenStudy (amistre64):

the chain rule applies when you have to differentiate a function

OpenStudy (amistre64):

its more noticable in nested functions tho

OpenStudy (amistre64):

\[[f(g(h(x)))]=f'(g(h(x)))*g'(h(x))*h'(x)*x'\]

OpenStudy (anonymous):

yes but how does one know when to differentiate an equation using this method when one is given an equation like \[f(x)=x^2 lnx\] what gives you the clue that this equation requires the chain rule

OpenStudy (anonymous):

?

OpenStudy (amistre64):

the ln is a function

OpenStudy (amistre64):

even tho its not a nested function perse. if the argument in it had been more complicated, it would show off better

OpenStudy (amistre64):

( )^2 is even a function, but its innards at the moment dont lend if to much showing off

OpenStudy (anonymous):

thank you

OpenStudy (amistre64):

(x^2+3x)^2, say we rename the innards: u(x) = x^2+3x this gives us something we can demonstrate better (u(x))^2 = 2(u(x)) * u'(x) is an application of the chain rule

OpenStudy (anonymous):

do you do statistics as well?

OpenStudy (amistre64):

only 13% of the time to remain relevant :)

OpenStudy (amistre64):

that joke is funny to a stat person :/

OpenStudy (anonymous):

lol my bad

OpenStudy (amistre64):

i remember alot from my elem stats class; but we never really got into poisson and such

OpenStudy (anonymous):

i always hated that still do.....can i ask you a few questions?

OpenStudy (amistre64):

sure

OpenStudy (amistre64):

its either a long question; or your running on internet explorer

OpenStudy (anonymous):

my problem is people always just tell us the answers but fail to tell us the explanation...so can you please right the explanation or rather the working out of this sum....

OpenStudy (anonymous):

long question lol .....and i am a slow typer lol

OpenStudy (amistre64):

i never "just tell the answer"; usually since i dont know the answer till we work it out to the end :)

OpenStudy (anonymous):

minoxidil was investigated (as a remedy for male baldness) in a large, 27-centre study where patients were randomly assigned to receive topical minoxidil or an identical-appearing placebo.ignore the centre=to-centre variation,suppose the preliminary results were as follows

OpenStudy (amistre64):

looks like its gona be a hypothesis testing

OpenStudy (anonymous):

the sample size for the minoxidil group was 310 and the sample size for the placebo was 309

OpenStudy (anonymous):

% with new hair growth for the minoxidil was 32 and the placebo was 20

OpenStudy (anonymous):

the question is .... CONSTRUCT A 95% CONFIDENCE INTERVAL FOR THE DIFFERENCE IN THE PROPORTIONS OF PATIENTS WITH NEW HAIR GROWTH FOR THE TWO GROUPS USING THE GIVEN DATA

OpenStudy (anonymous):

THE OTHER QUESTION IS ............... use the data to test Ho:p1-p2=0 versus H1:p1-p2 is not equal to 0....give the p-value for the test

OpenStudy (amistre64):

2 hypothesis testing .... and the first one resembles a difference of the means .... both are a bit out of my comfort zone, abd i would have to rely on something like stattrek to keep me on track

OpenStudy (amistre64):

the formulas are reminescent of single stuff; but just adding the data of the 2 of them in some fashion ....

OpenStudy (anonymous):

lol so it is a two tailed test?

OpenStudy (amistre64):

pfft, i cant even determine that from the content :)

OpenStudy (amistre64):

the difference of proportions i believe is a confidence interval for the difference of the means

OpenStudy (anonymous):

wow so how would you go abt answering this question

OpenStudy (amistre64):

http://stattrek.com/hypothesis-test/difference-in-proportions.aspx id direct someone to this for and talk them thru things they might have difficulties with; but the explanations on the site are pretty cogent

OpenStudy (anonymous):

i would just divide the confidence level by 2 than find the rejection area buth the statistic test wud jst confuse me

OpenStudy (amistre64):

we have a few stats savants about from time to time; zarkon, satellite, and james come to mind. There are others as well, but they tend to shy away for some odd reason

OpenStudy (anonymous):

lol ok thank you.......i'l contact them right away

OpenStudy (amistre64):

i wish i could have been more help on that one, but we never really got past the confidence intervals of a single proportion :/ good luck with it tho ;)

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