Question

I need some work checked over! :) I fear that when taking the following derivative(s), I goofed up, and just want a second mind to do a once over of this math.

So, the original functions is "y = (x^4 + 1)/x^2".
I got the following for the first and second derivative respectively:
"(2x^5 + 2x)/x^4",
"2 + ((10x^4)/x^8)"

Any and all help is greatly appreciated! :)

Answer 1

mmm, not correct...

Answer 2

Neither of them? My calc teacher actually gave me the first one, but I found the second one on my own... I hope my Calc teacher knows what he's doing ;)

Answer 3

\[y=\frac{x^4+1}{x^2}\]right?

Answer 4

Yep!

Answer 5

@wolfe8

Answer 6

Well, we can break that into two fractions:
\[y = \frac{x^4}{x^2} + \frac{1}{x^2}\]right?

Answer 7

Yeah, I suppose. Yes, it can be. Is that a better way to do it, rather than using the Quotient Rule in terms of the whole function, right away?

Answer 8

We can rewrite that as \[y = x^{4-2} + x^{-2} = x^2 + x^{-2}\]
What's the derivative of that? \[\frac{dy}{dx} = 2x^1 + (-2)x^{-2-1} = 2x-2x^{-3}\]

Answer 9

Similarly, for the second derivative, we'll have
\[\frac{d^2y}{dx^2} = 2-2*(-3)x^{-3-1} = 2+6x^{-4}\]

Answer 10

Yes... That totally makes sense to me, but... Why didn't my teacher use that method, and if they're both legitimately both algebraically right... then... What one is "Right"? I see how what you did works, and it's so much easier! But... Then why have the Quotient Rule? :/

Answer 11

@agent0smith

Answer 12

Well, if we had something like
\[y = \frac{x^4}{x^2+1}\]we couldn't have done what we did here.

Answer 13

True...

Answer 14

One good argument for not using the quotient rule for this problem is that the people who tried it apparently screwed up :-)

Answer 15

This could be expressed in many ways such as;
\[2x-2/x ^{3}\]or\[2x ^{4}-2 \over x ^{3}\] etc.

Answer 16

why don't you restate the quotient rule for me, and tell me what you plugged in for the various variables? Maybe we can find out where you went off the tracks...

Answer 17

However, none would be what the calc teacher did.

Answer 18

y' = (u'v - uv')/v^2

My u = (x^4 + 1) ...for finding y" u = (2x^5 + 2x)
My v = x^2 ... for finding y" v = x^4

Answer 19

And yeah, I get the rearrangements there @radar. :)

Answer 20

Are those accurate ?

Answer 21

\[u = x^4 + 1\]\[u' = 4x^3\]\[v = x^2\]\[v' = 2x\]\[y' = \frac{u'v - uv'}{v^2} = \frac{4x^3*x^2-(x^4+1)*2x}{(x^2)^2} = \frac{4x^5-2x^5-2x}{x^4} = \frac{2x^5-2x}{x^4}\]\[=2x-2x^{-3}\]

Answer 22

So I guess what I'm really looking to ascertain is when/ how do I know when to use one method vs. another, and what determines that, and how many ways are there ? What, concretely, and as a matter of fact, makes one way " Right" and another "wrong" in situations like this?? :/

Answer 23

actually I guess the 1st one was correct, I just didn't simplify it.