Easy second derivative question

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The discussion revolves around finding the second derivative of the equation y^2 - 3x = 7. The first derivative was calculated as 3/(2y), but the user struggled with the second derivative, initially arriving at an incorrect expression. After guidance, the correct approach involves applying the chain rule and substituting the first derivative back into the second derivative calculation. The final expression for the second derivative is clarified as -3/(2y^2) multiplied by the first derivative. The user expresses gratitude for the assistance in resolving their confusion.
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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 \frac{dy}{dx}=\frac{3}{2y}=\frac{3}{2}y^{-1}
diff. w.r.t x again

\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{dy}{dx}

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

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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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