Question

find the second derivative of (4-2x^2)^10

Answer 1

The answer is apparently 80(4-2x^2)(19x^2-2) and I am NOT getting it...

Answer 2

y' = 10(4-2x^2)^9 * (-4x)

Answer 3

did you get that

Answer 4

Yep! I got that exactly!

Answer 5

\[\frac{d}{dx}(4-2x^{2})^{10} = 10*(4-2x^{2})^{9}*(-4x) = -40x*(4-2x^{2})^{9}\]
// chain rule
\[\frac{d}{dx}(-40x*(4-2x^{2})^{9}) = -40*(4-2x^{2})^{9} + (-40x*9*(4-2x^{2})^{8}*(-4x))\]

\[= [-40*(4-2x^{2})^{8}]*[(4-2x^{2})+9*(x*-4x)]\]
\[=[ -40*(4-2x^{2})^{8}]*[(4-2x^{2}-36x^{2}]\]
\[= [ -40*(4-2x^{2})^{8}]*[(4-38x^{2})]\]

Answer 6

next use the product rule
so
y'' = 10(4-2x^2)*(-4) + (10(-4x))(-4x)

Answer 7

sorry hold on i forgot the x^9

Answer 8

y' = 10(4-2x^2)^9 * (-4x)
y'' = (10(4-2x^2)^9) * (-4) + (10*8(4-2x^2)^8)*-4x*-4x

Answer 9

then simplify

Answer 10

AHA! Thanks soooo much! I'm reviewing right now, so I'm trying to troubleshoot.

Answer 11

your welcome, guess @tomo already had it covered for ya though

Answer 12

Okay wait a sec... where'd the two -4x's come from?

Answer 13

from takin the second half of the product rule
y' = 10(4-2x^2)^9 * (-4x)
when you do d/dx(10(4-2x^2)^9) * -4x
in the d/dx you have another chain rule

Answer 14

wouldnt it be -4x(y'(10(4-2x^2)^9)+10(4-2x^2)^9(y'(-4x)

Answer 15

well the first half of the product rule the way you are doing it.

Answer 16

-4x(y'(10(4-2x^2)^9)
this part d/dx(10(4-2x^2)^9) you have to use a chain rule and so another -4x pops out

Answer 17

Ack! I thought I missed one of the little buggers! Okay, I've got it now! Thanks again for the help!

Answer 18

yep just simplify it and make sure its right. its easy to mess up these things

Answer 19

that may not be right and i dont think the answer you said it was was right either. the correct answer is
40 (4-2 x^2)^8 (38 x^2-4)

Answer 20

thats what i got using wolfram alpha

Answer 21

The answers I'm given are never on par with wolfram alpha... would yo mind would checking me here? This is from using -40x(4-2x^2)^9
(-40x(9(4-2x^2)^8(-4x))+(10(4-2x^2)^9*(-40)

Answer 22

did you forget a 10 in the first part

Answer 23

since there is one in the second

Answer 24

so then:
(-40x(90(4-2x^2)^8(-4x))+(10(4-2x^2)^9*(-40)

Answer 25

caught a mistake in my answer
y'' = (10(4-2x^2)^9) * (-4) + (10*8(4-2x^2)^8)*-4x*-4x
shoulb be
y'' = (10(4-2x^2)^9) * (-4) + (10*9(4-2x^2)^8)*-4x*-4x
^

Answer 26

So my answer is correct?

Answer 27

no

Answer 28

Ugh... I hate this stuff. What am I doing wrong?!

Answer 29

one step at a time
y' = 10(4-2x^2)^9 * (-4x)
product rule (d/dx)1st * 2nd = d/dx 2nd *first
for d/dx 1st*2nd
d/dx(10(4-2x^2)^9) = 9*10(4-2x^2)^8*(-4x) * -4x
second

for d/dx 2nd * 1st

d/dx(-4x) = -4 so te wholpart is -4(10(4-2x^2)^9)

then add the two parts

9*10(4-2x^2)^8*(-4x) * -4x -4(10(4-2x^2)^9)

Answer 30

AHA! Okay I've got it now! Thanks again for all the help!

Answer 31

no prob man