Question

Ok guys this is a derivatives question. I just want to make sure I am on the right path--
y=(x^2+1)^2 + 3(x^2-1)

Answer 1

Did you make an attempt at the problem? What did you come up with? :)

Answer 2

lol. I was making an attempt and then I got confused. I started off by using the power rule for the two set I am multiplying by. I got stuck on the second set 3(x^2-1)^2. I'm not sure if I am supose to partly use the constant rule with that one. I know in the end I will have to utilize the sum rule on all.

Answer 3

The second set just really confused me a bit

Answer 4

\[\large y=(x^2+1)^2+3(x^2-1)\]Applying the power rule to the first part, we take the derivative of the `outermost` function, that being the square on the brackets, then we'll apply the chain rule.\[\large \rightarrow \qquad 2(x^2+1)^{2-1}\]Then the chain rule tells us to multiply by the derivative of the inside,\[\large 2(x^2+1)^{1}\frac{d}{dx}(x^2+1)\]The fancy derivative operator is to show that we need to take the derivative of that part still. Taking the derivative gives us,\[\large 2(x^2+1)(2x)\]

This is how the first one works out. Were you able to get to this? c: The next one will work out very similarly. I guess it's just the 3 that's confusing you?

Answer 5

yes. I definitely got that for the first set. I am just totally confused on the second. I made a mistake the second set is square so it is: 3(x^2-1)^2. I was wandering if I would do the constant or the power rule first

Answer 6

When taking derivatives, completely ignore `constant coefficients`. They will have no effect on taking the derivative.\[\large \color{royalblue}{\frac{d}{dx}}3(x^2-1)^2 \qquad = \qquad 3\color{royalblue}{\frac{d}{dx}}(x^2-1)^2\]See how I pulled the 3 outside of the derivative operator? We can ignore it, and do the same steps we did for the first part.

Answer 7

oh ok. I will do that now. thanks

Answer 8

so you would get [3(2)(x^2-1)(2x)\]

Answer 9

please ignore the slash at the end. typo

Answer 10

to simplify it some more 6(x^2-1)(2x)

Answer 11

So the derivative of the inside gave you 2x again?
Yah looks good! :)

Answer 12

To make it look a little cleaner, do a tiny bit more simplification on both terms!\[\large 6(x^2-1)(2x) \qquad = \qquad 12x(x^2-1)\]

Answer 13

I was just going to ask you if I should do that. The other side simplified is 4x(x^2-1)

Answer 14

Yah looks good.

Answer 15

Ok. I have an answer. I'm not quite sure it is correct though. I added the two sides by taking the derivatives of each side then adding. So I got 4(2x)+12(2x). Then I simplified it to 8x+24x--->32x...I might be way off

Answer 16

Should I not have simplified it further and left it as is?

Answer 17

\[\large y=(x^2+1)^2+3(x^2-1)^2\]
\[\large y'=2(x^2+1)(2x)+3(x^2-1)(2x)\]\[\large y'=4x(x^2+1)+6x(x^2-1)\]You should end up with this, hmm.

Answer 18

ok. I understand what you did. But why didn't you take the constant 3 from the beginning of the equation into consideration? That is what I did so I got
4x(x^2-1)+12x(x^2-1). Then I applied the sum rule and added it up. Pretty much I took the derivative of each again then added: 4(2x)+12(2x)--->8x+24x--->32x?

Answer 19

I might be completely wrong

Answer 20

Oh yes I missed a factor of 2 in there, sorry bout that :)\[\large y'=4x(x^2+1)+12x(x^2-1)\]
Took the derivative again? Were you looking for the second derivative of y?

Answer 21

lol...I don't know. Just looking at the instructions now it says Differentiate. Maybe that's why I did it. I thought that possibly meant simplify it. As you can see calculus is not my thing

Answer 22

So is this what the instructions look like? :o

Differentiate this function:
\[\large y=(x^2+1)^2 + 3(x^2-1)^2\]

Answer 23

lol...Exactly. You're so smart!

Answer 24

lol

Answer 25

Taking the derivative gave us this,
\[\large y'=4x(x^2+1)+12x(x^2-1)\]

I'm not really sure what you did beyond that though.
Sum Rule? :O hmm

With the way it's written right now, it looks like the insides are not the same, so these can't be added together.\[\large y'=4x(\color{orangered}{x^2+1})+12x(\color{royalblue}{x^2-1})\]See how the orange and blue are different? We can't combine them.
Is that what you were trying to do? <:o

Answer 26

lol..ok this is the sum rule: d/dx[f(x)+g(x)]=d/dx[f(x)+d/dx[g(x)]

Answer 27

I just applied the rule to both sides d/dx 4x(x^2-1)+d/dx 12x(x^2-1)

Answer 28

which gave: 4(2x)+12(2x)--->8x+24x--->32x? Like I said I maybe off with this one. Hopefully I did not confuse you

Answer 29

So the sum rule is telling us that THIS:
\[\large \frac{d}{dx}\left[(x^2+1)^2+3(x^2-1)^2\right]\]

Is equal to THIS:\[\large \frac{d}{dx}(x^2+1)^2+\frac{d}{dx}3(x^2-1)^2\]

Then taking the derivative of each term separately gives us what we had before. It looks like you're just applying this Sum Rule at the end by mistake, instead of at the beginning.

Answer 30

ok. gotcha. thanks so much!