bmelyk
need someone to walk me through this question, please:



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bmelyk
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Use the appropriate rule or combination of rules to find the derivative of the function defined below.
\[y=2(\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 }\]

bmelyk
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not even sure how to start this one. ;(
looks like it would the the chain rule of course.

hartnn
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first what is derivative of x^{1/2} = ?

bmelyk
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1/2x^1/2

hartnn
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we always go from outer function to inner function

hartnn
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\(y'=2(1/2)(\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 }d/dx(\sin(6x)+\cos(x^3))\)

bmelyk
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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
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u got my last comment ? how i got that ?

bmelyk
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and i had that so far.

bmelyk
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yesi got that.

hartnn
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so what is d/dx(sin(6x)+cos(x^3)) ?

bmelyk
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do i use the product rule for sin*6x and cos*x^3?

hartnn
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there is no product, so NO
u can treat them individually also
what is d/dx (sin 6x) = ?

bmelyk
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cos6x

bmelyk
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sinx^3

hartnn
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only ?

hartnn
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i mean only cos 6x ?

bmelyk
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times 6?

hartnn
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u need chain rule here too, right ?

hartnn
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yes! times 6, because u'll have d/dx(6x)

bmelyk
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cox6x*6

bmelyk
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cos6x*6

bmelyk
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????

hartnn
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yup, what about 2nd term ?
d/dx (cos (x^3)) = ?

bmelyk
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sinx^3(3x^2)

hartnn
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absolutely correct!
u got it :)

hartnn
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so which option is it ?

bmelyk
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e??

hartnn
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put everything together, and u'll get it.

hartnn
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yes, correct :)

bmelyk
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awesome!

hartnn
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i hope chain rule is clearer to u now.....