Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

need someone to walk me through this question, please:

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=2(\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 }\]
not even sure how to start this one. ;( looks like it would the the chain rule of course.
first what is derivative of x^{1/2} = ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

1/2x^-1/2
we always go from outer function to inner function
\(y'=2(1/2)(\sin(6x)+\cos(x^3))^\frac{ -1 }{ 2 }d/dx(\sin(6x)+\cos(x^3))\)
options for the right answer are: \[a)2*\frac{ \cos(6x)-\sin(x^3) }{ (\sin(6x)+cox(x^3))^1/2 }\] \[b)\frac{ \cos(6x) +\sin(x^3)}{ (\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 } }\] \[c)\frac{ 12 }{ 6x^\frac{ 1 }{ 2 } }*\cos(6x)-\frac{ 6 }{ \cos(x^3)^\frac{ 1 }{ 2 } }*\sin(x^3)*x^2\] \[d)\frac{ 1 }{ (\sin(6x)^\frac{ 1 }{ 2 }+\cos(x^3))^\frac{ 1 }{ 2 } }\] \[e)\frac{ 6*\cos(6x)-3x^2*\sin(x^3) }{ (\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 } }\] f) same as e only 2 outside of the fraction. \[g) \frac{ 2 }{ \sin(6x)+\cos(x^3))^1/2 }\] \[h) \frac{ 6 }{ \sin(6x)^\frac{ 1 }{ 2 } }*\cos6x-\frac{ 3 }{ \cos(x^3)^\frac{ 1 }{ 2 } }*\sin(x^3)*x^2\]
u got my last comment ? how i got that ?
and i had that so far.
yesi got that.
so what is d/dx(sin(6x)+cos(x^3)) ?
do i use the product rule for sin*6x and cos*x^3?
there is no product, so NO u can treat them individually also what is d/dx (sin 6x) = ?
cos6x
-sinx^3
only ?
i mean only cos 6x ?
times 6?
u need chain rule here too, right ?
yes! times 6, because u'll have d/dx(6x)
cox6x*6
cos6x*6
????
yup, what about 2nd term ? d/dx (cos (x^3)) = ?
-sinx^3(3x^2)
absolutely correct! u got it :)
so which option is it ?
e??
put everything together, and u'll get it.
yes, correct :)
awesome!
i hope chain rule is clearer to u now.....

Not the answer you are looking for?

Search for more explanations.

Ask your own question