A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
how do u solve a problem that looks like this:
y=cubic root of sin sqx
anonymous
 5 years ago
how do u solve a problem that looks like this: y=cubic root of sin sqx

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you could start by writing it out in a usable manner :) ....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is written in a usable manner, im just confused because professor didnt give us any instruction in what to do

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what does this mean: sin sqx?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0aha!! now it looks like something I can use :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 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
 5 years ago
Best ResponseYou've already chosen the best response.0The "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
 5 years ago
Best ResponseYou've already chosen the best response.0whay=what...... my keyboard hates me..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank 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
 5 years ago
Best ResponseYou've already chosen the best response.0chain rule is good f(g(h(x))) right? f(x) = cbrt(x); g(x) = x^2; h(x) = sin(x)...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, the problem is i dont know how to get the derivative of cbrt sin^2x.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin^2(x) can be a clumsy notation; it is just a way of writing (sin(x))^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0derivative eh..you trying to derive this equation? that would have been helpful to know :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awww sorry, i just found out i had to to that.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0rework the equation into exponents; cbrt = ^(1/3) and sqrt = ^(1/2) when exponents have exponents; they multiply together..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin^2^(1/2)^(1/3) = sin^(1/3); think of this as x^(1/3) and derive

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I get: (1/3)cos(x) ^(2/3): which can be cleaned up some

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01  3 cbrt(cos^2(x)) Whatcha think about that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my friend got almost the answer, thank you so much!!!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0almost the answer? I wonder what an "almost the answer" looks like :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he got a 2 instead of 1.. but we are all confused so i'll go with ur answer, thank you

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0check my work, by all means, after all..im an idiot in disguise :) yw!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lolz, yeah im copying what you tyoed and i'm going to check it .. thank you tho
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.