Easy second derivative question

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Homework Help Overview

The problem involves finding the second derivative of the equation y² - 3x = 7. Participants are discussing the process of differentiation and the application of the chain rule in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the second derivative after calculating the first derivative, expressing confusion about their result not matching the provided answer choices. Other participants discuss the application of the chain rule and substitution of the first derivative into the second derivative calculation.

Discussion Status

Participants are actively engaging with the problem, providing insights into the differentiation process. Some guidance has been offered regarding the proper substitution of the first derivative into the second derivative calculation, indicating a productive direction in the discussion.

Contextual Notes

The original poster expresses uncertainty about their calculations and mentions a pattern of errors in their recent work, which may affect their confidence in solving similar problems.

demersal
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[SOLVED] easy second derivative question

Homework Statement


"If y^2 - 3x =7 what is the second derivative?"

Homework Equations



Answer choices:

A) -6/7y^3
B) -3/y^3
C) 3
D) 3/2y
E) -9/4y^3


The Attempt at a Solution



I got the first derivative to be: 3/2y
Second derivative: -6/(4y^2)

I cannot for the life of me figure out what I did wrong, but I know it's something since that's not a choice. I've been messing up a lot lately, so I think I'm making some sort of vital error. A step by step explanation would be appreciated!

You guys are so awesome here, thank you for all that you do.
 
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If y'=3/(2y)=(3/2)*y^(-1) (and it is) then y''=(3/2)*(-1)*y^(-2)*y'. Notice the y' on the right side coming from the chain rule. Now substitute your previous expression for y' for that y'.
 
well the first derivative is [itex]\frac{dy}{dx}=\frac{3}{2y}=\frac{3}{2}y^{-1}[/itex]
diff. w.r.t x again

[tex]\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{dy}{dx}[/tex]

which should give...[tex]\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{3}{2}y^{-1}[/tex]

I think you multiplied wrongly
 
Dick, you are beyond wonderful. I kept messing up because I wasn't subbing for y' properly, but now I see the trick to use the first derivative.

THANK YOU!
 

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