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July 6th, 2014, 02:01 PM   #1
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problem with simplification when taking the derivative of a derivative

edit... solved it by just working through it slowly....

So, first off ty to wheoever reads this, this is my first post so I didn't know whether to put this in the algebra forum, the calculus forum or some other forum.


I'm getting confused in some of simplification in one of my calculus homework problems.

I'm meant to get d''/x'' of
15x + y^2 = 2

First taking the derivative I get y'= -15x/y, and so far so good.

When plugging the derivative in the second time using the quotient rule and plugging in -15x/y for y' it becomes


-15y - (-15x)(-15x/y)
---------------------
y^2


This somehow simplifies to

-15(y^2+15x^2)
-----------------
y^3


The book skips the simplification in going through the answer so I don't know the steps they took to simplify.

I thought the y in -15x/y would only go to the bottom of the fraction so how does -15y become -15y^2?

or did the book multiply both top and bottom by y to get rid of the y in -15x/y, but in that case wouldnt the x take on a y as well and we'd get

-15(y^2 + 15x^2 * y)
--------------------
y^3

Last edited by lackofimagination; July 6th, 2014 at 02:08 PM.
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July 6th, 2014, 08:05 PM   #2
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Quote:
Originally Posted by lackofimagination View Post



or did the book multiply both top and bottom by y to get rid of the y in -15x/y, but in that case wouldnt the x take on a y as well and we'd get

-15(y^2 + 15x^2 * y)
--------------------
y^3
No, look at this:

(-15x)(-15x/y)*y =

(-15)(-15)(x)(x/y)*y =

+ $\displaystyle (15)(15)(x)\dfrac{x}{y}*y \ = $

+ $\displaystyle (15)(15)x^2$

----------------------------------------------------

In general, with three numbers, a, b, and c,

(a)(b)(c) = (a*b)(c) or (a*c)(b) or (b*c)(a).


But, for instance, (a)(b)(c) $\displaystyle \ \ne \ (a*b)(a*c).$
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