Finding derivative of a function that is to the third under a square root

srcharme

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Joined
Feb 24, 2015
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Hello,

I'm new here and I'm not really sure if I'm doing this correctly, so I apologize if I make any horrible posting errors.
I have to find the derivative of y=sqrt((3x2+2)3)

I'm unsure about the order that I need to apply the square root/exponent.
Here is what I have:
y'=((3x2+2)3)1/2
=(3x2+2)1.5
=1.5(3x2+2)0.5(6x)

Is this correct?
Any help would be appreciated, thank you!
 
Hello,

I'm new here and I'm not really sure if I'm doing this correctly, so I apologize if I make any horrible posting errors.
I have to find the derivative of y=sqrt((3x2+2)3)

I'm unsure about the order that I need to apply the square root/exponent.
Here is what I have:
y'=((3x2+2)3)1/2
=(3x2+2)1.5
=1.5(3x2+2)0.5(6x)

Is this correct?
Any help would be appreciated, thank you!
Looks good. Actually, they way you did it was probably best but it doesn't make any difference how you proceed. Letting
f(x) = 3x2+2
you have
((f3)1/2)' = (1/2) ((f3)-1/2) 3 f2 f' = 1.5 f-3/2 f4/2 (6x) = 1.5 f1/2 (6x) = 1.5 (3x2+2)0.5 (6x)
 
I have to find the derivative of y=sqrt((3x2+2)3)
I'm unsure about the order that I need to apply the square root/exponent.
Here is what I have: y'=((3x2+2)3)1/2=(3x2+2)1.5
In some circles it is considered bad form to use a decimal as an exponent.
 
Hello,

I'm new here and I'm not really sure if I'm doing this correctly, so I apologize if I make any horrible posting errors.
I have to find the derivative of y=sqrt((3x2+2)3)

I'm unsure about the order that I need to apply the square root/exponent.
Here is what I have:
y'=((3x2+2)3)1/2
=(3x2+2)1.5
=1.5(3x2+2)0.5(6x)

Is this correct?
Any help would be appreciated, thank you!
You really should multiply the 1.5 and 6x to get 9x.
My biggest concern is that you wrote y'=((3x2+2)3)1/2 which is NOT true! It is y (not y') that equals ((3x2+2)3)1/2
 
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