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Mathematics 15 Online
OpenStudy (anonymous):

how do u solve a problem that looks like this: y=cubic root of sin sqx

OpenStudy (amistre64):

you could start by writing it out in a usable manner :) ....

OpenStudy (anonymous):

it is written in a usable manner, im just confused because professor didnt give us any instruction in what to do

OpenStudy (amistre64):

what does this mean: sin sqx?

OpenStudy (anonymous):

sin^2 X, sorry :D

OpenStudy (amistre64):

aha!! now it looks like something I can use :)

OpenStudy (amistre64):

y = cbrt(sin^2(x)) The "solution" will be all legal "x" values that make the statement true. It most likely is the graph of the equation.

OpenStudy (amistre64):

The "Domain" is whay we use apply a value to "x"; and as long as we only use what is allowed by the Domain, the equation pops out a solution.

OpenStudy (amistre64):

whay=what...... my keyboard hates me..

OpenStudy (anonymous):

thank you, i think i have to use the chain or power rule somehow, but i have problems solving it, bacuase of the sin^2x

OpenStudy (amistre64):

chain rule is good f(g(h(x))) right? f(x) = cbrt(x); g(x) = x^2; h(x) = sin(x)...

OpenStudy (anonymous):

yeah, the problem is i dont know how to get the derivative of cbrt sin^2x.

OpenStudy (amistre64):

sin^2(x) can be a clumsy notation; it is just a way of writing (sin(x))^2

OpenStudy (amistre64):

derivative eh..you trying to derive this equation? that would have been helpful to know :)

OpenStudy (anonymous):

awww sorry, i just found out i had to to that.

OpenStudy (amistre64):

rework the equation into exponents; cbrt = ^(1/3) and sqrt = ^(1/2) when exponents have exponents; they multiply together..

OpenStudy (amistre64):

sin^2^(1/2)^(1/3) = sin^(1/3); think of this as x^(1/3) and derive

OpenStudy (amistre64):

I get: (1/3)cos(x) ^(-2/3): which can be cleaned up some

OpenStudy (amistre64):

1 ------------- 3 cbrt(cos^2(x)) Whatcha think about that?

OpenStudy (anonymous):

my friend got almost the answer, thank you so much!!!!

OpenStudy (amistre64):

almost the answer? I wonder what an "almost the answer" looks like :)

OpenStudy (anonymous):

he got a 2 instead of 1.. but we are all confused so i'll go with ur answer, thank you

OpenStudy (amistre64):

check my work, by all means, after all..im an idiot in disguise :) yw!

OpenStudy (anonymous):

lolz, yeah im copying what you tyoed and i'm going to check it .. thank you tho

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