anonymous
  • anonymous
how do u solve a problem that looks like this: y=cubic root of sin sqx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
you could start by writing it out in a usable manner :) ....
anonymous
  • anonymous
it is written in a usable manner, im just confused because professor didnt give us any instruction in what to do
amistre64
  • amistre64
what does this mean: sin sqx?

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anonymous
  • anonymous
sin^2 X, sorry :D
amistre64
  • amistre64
aha!! now it looks like something I can use :)
amistre64
  • 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.
amistre64
  • 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.
amistre64
  • amistre64
whay=what...... my keyboard hates me..
anonymous
  • 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
amistre64
  • amistre64
chain rule is good f(g(h(x))) right? f(x) = cbrt(x); g(x) = x^2; h(x) = sin(x)...
anonymous
  • anonymous
yeah, the problem is i dont know how to get the derivative of cbrt sin^2x.
amistre64
  • amistre64
sin^2(x) can be a clumsy notation; it is just a way of writing (sin(x))^2
amistre64
  • amistre64
derivative eh..you trying to derive this equation? that would have been helpful to know :)
anonymous
  • anonymous
awww sorry, i just found out i had to to that.
amistre64
  • amistre64
rework the equation into exponents; cbrt = ^(1/3) and sqrt = ^(1/2) when exponents have exponents; they multiply together..
amistre64
  • amistre64
sin^2^(1/2)^(1/3) = sin^(1/3); think of this as x^(1/3) and derive
amistre64
  • amistre64
I get: (1/3)cos(x) ^(-2/3): which can be cleaned up some
amistre64
  • amistre64
1 ------------- 3 cbrt(cos^2(x)) Whatcha think about that?
anonymous
  • anonymous
my friend got almost the answer, thank you so much!!!!
amistre64
  • amistre64
almost the answer? I wonder what an "almost the answer" looks like :)
anonymous
  • anonymous
he got a 2 instead of 1.. but we are all confused so i'll go with ur answer, thank you
amistre64
  • amistre64
check my work, by all means, after all..im an idiot in disguise :) yw!
anonymous
  • anonymous
lolz, yeah im copying what you tyoed and i'm going to check it .. thank you tho

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