anonymous
  • anonymous
need someone to walk me through this question, please:
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Use the appropriate rule or combination of rules to find the derivative of the function defined below. \[y=2(\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 }\]
anonymous
  • anonymous
not even sure how to start this one. ;( looks like it would the the chain rule of course.
hartnn
  • hartnn
first what is derivative of x^{1/2} = ?

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

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