Question

Can someone help this?

The derivative of y = (-x)/(1-x^2)^0.5 is ?

Answer 1

do you have the selected answer?

Answer 2

love derivatives. so you are using a couple different things here (also if you want a very nice step by step try wolframalpha, i get sloppy sometimes). first you need to use the quotient rule which is:

(f(x)/g(x))'= [f'(x)g(x)-g'(x)f(x)]/g(x)^2

so knowing that we are eventually going to be putting it in that form, lets first find the derivative of the top. this one isn't too bad, since the derivative of x is 1. so factoring in the - sign we get:

-x'=-1

now lets look at the bottom part. since this is a little more complicated, we are going to use the chain rule. so we need to take the derivative of the outside times the derivative of the inside:

(1-x^2)^(1/2)'=1/2{[1-x^2]^(-1/2)}(-2x)

reducing that we get:

-x/(1-x^2)^(1/2)

now lets put it all together:

-(1-x^2)^(1/2)-(-x)(-x/(1-x^2)^(1/2)) / 1-x^2

Answer 3

i feel like i should go to a book signing now.

Answer 4

hmm my answer is [-1+(x^2)(1-x^2)^-(3/2) ]

Answer 5

true story, forgot the exponent on the bottom was negative. uber fail

Answer 6

actually also wrong, i just wolframed that pellet. it gave me:

-1/(1-x^2)^(3/2)

Answer 7

try to cancel the denominator by factoring out the 1-x^2

Answer 8

ah ok good

Answer 9

i got (-1-x^2)/(1-x^2)^(3/2)

Answer 10

you are very close try again..continue solving til you get y=-1/(1-x^2)^(3/2)

Answer 11

can you show me how to get it?i still cant get it.

Answer 12

ryt, did you use the derivative of the quotient formula? use that

Answer 13

d(u/v)=(vdu-udv)/v^2

Answer 14

yea i used it but cant get that answer.

Answer 15

why is it 1 -2x?
i thought it will be -2x ?

Answer 16

dy/dx = d[(-x)/(1-x^2)^0.5]/dx
=((1-x^2)^1/2)(-1)-(-x)(1/2)(1-x^2)^-1/2(-2x)]/(1-x^2)

Answer 17

yes -2x, sorry wrong typing lol...

Answer 18

oh okay. so the answer is (-1-x^2)/(1-x^2) ?

Answer 19

=(-(1-x^2)^1/2)-(x^2)((1-x^2)^-1/2]/(1-x^2)

Answer 20

no the answer is dy/dx =-1/(1-x^2)^(3/2)

Answer 21

how you get that from the previous working?

Answer 22

=(-(1-x^2)^1/2)-(x^2)((1-x^2)^-1/2]/(1-x^2) , factoring out (1-x^2)
=(1-x^2)[-(1-x^2)^1/2 -x^2 (1-x^2)^-3/2]/(1-x^2)
=[-(1-x^2)^1/2 -x^2 (1-x^2)^-3/2]
=-1/(1-x^2) -x^2 /(1-x^2)^3/2
=-(1-x^2)-x^2 /((1-x^2)^3/2 thats it..
=-1/(1-x^2)^3/2 ans

Answer 23

did you get it ryt?

Answer 24

how to do you get the answer from the pervious one?(5th and 6th line of working)

Answer 25

hi ryt sorry my pc hanged up, i didnt get your question on time

Answer 26

oh its okay:) i get it :D thanks!!