Question

Can anyone help me find me error on my work? (Derivatives)

Answer 1

I'll be writing out step by step so that it is easier to find out.

Answer 2

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

Answer 3

\[y'=\frac{ (-9x^2+3)(\sqrt{2-x^2}-(-3x^3+3x)(\frac{ 1 }{ 2 })(2-x^2)^\frac{ -1 }{ 2 }(-2x) }{ (\sqrt{2-x^2)^2} }\]

Answer 4

the power of 2 is outside the square root. My mistake.

Answer 5

correct :)

Answer 6

ok from now on I will simplify the numerator part only, so I will use the denominator later.

Answer 7

\[y'= -9x^2\sqrt{2-x^2}+3\sqrt{2-x^2}-\frac{ (-3x^3+3x)(-2x)}{ 2\sqrt{2-x^2} }(-3x^3+3x)(-2x)\]

Answer 8

delete the part after the fraction. That wasn't suppose to be there.

Answer 9

correct:)

Answer 10

\[y'=-9x^2\sqrt{2-x^2}+3x \sqrt{2-x^2}-\frac{ 3x^4+3x^2 }{ \sqrt{2-x^2} }\]

Answer 11

I'm starting to simplify it as much as I can by finding common denominators and so on.

Answer 12

\[y'= -9x^2\sqrt{2-x^2}+\frac{ 3(2-x) }{ \sqrt{2-x^2} }- \frac{ 3x^4+3x^2 }{ \sqrt{2-x^2} }\]

Answer 13

\[y'= -9x^2\sqrt{2-x^2}+\frac{ 6-3x^2-3x^4-3x^2 }{ \sqrt{2-x^2} }\]

Answer 14

\[y'= -9x^2\sqrt{2-x^2} + \frac{ -3x^4 -6x^2 +6 }{ \sqrt{2-x^2} }\]

Answer 15

wait

Answer 16

\[y'= \frac{ -9x^(2-x^2) }{ \sqrt{2-x^2} }-\frac{ 3x^4+6x^2-6 }{ \sqrt{2-x^2} }\]

Answer 17

what's the equation after i said correct

Answer 18

Derivative without using the denominator.

Answer 19

wait

Answer 20

I think I just found out where is the problem

Answer 21

\[-\frac{(-3x^4 -3x^2) }{ \sqrt{2-x^2} }\]

Answer 22

this is the answer
i solved it in my book