Question

Find the derivative of y=(x^2 +1) sqrt(2-x^2)

Answer 1

\[y' = 2x \sqrt{2-x^2} - \frac{ 2x^3 - 2x }{ 2\sqrt{2-x^2} }\]

Answer 2

use product rule

Answer 3

that's what I obtained from doing the product rule.

Answer 4

almost correct
except the sign in between 2x^3 - 2x will it not be
2x(x^2+1)

Answer 5

is (-2x)(x^2+1) though

Answer 6

yup

Answer 7

\[\Large\rm y' = 2x \sqrt{2-x^2} \color{red}{-} \frac{ 2x^3 - 2x }{ 2\sqrt{2-x^2} }\]This red negative sign means the same thing as,\[\Large\rm y' = 2x \sqrt{2-x^2} +\frac{ -(2x^3 - 2x) }{ 2\sqrt{2-x^2} }\]

Answer 8

ok

Answer 9

You wanted the second term to be negative, yes?
Your sign is just a little out of wack.

Answer 10

I see, so is - (2x^3 +2x)

Answer 11

Mmm yah that looks better :)

Answer 12

I am trying to simplify it further now. I found common denominator

Answer 13

I ended with \[y'= \frac{ -6x^3+6x }{ 2\sqrt{2-x^2} }\]

Answer 14

Mmm ok good, that's what I'm coming up with also.
Looks like you can go a tad further.
Too many 2's everywhere..

Answer 15

\[y'= \frac{ -3x^2 +3x }{ \sqrt{2-x^2} }\]

Answer 16

Yah there we go! :)

Answer 17

Oh and to find the domain of this. I just take out the 2-x^2 from the square root and solve it to zero as 2-x^2>0

Answer 18

so I end up having x<+- sqrt(2)

Answer 19

or [-sqrt(2),sqrt(2)]

Answer 20

\[\Large\rm 0=(x^2+1)\sqrt{2-x^2}\]Yes you apply your Zero-Factor Property thing:\[\Large\rm 0=\sqrt{2-x^2}\]And your x values look correct! yay!

Answer 21

We ignored the other factor since it's giving us all reals for the domain.

Answer 22

ok, thanks you! I will be posting later on a harder problem that I have no idea how to start =.=

Answer 23

Cool c:
Well we've got a ton of people here who enjoy the Calculus Problems.
So you shouldn't be short on finding assistance.

Answer 24

Thanks