Question

Need help finding the derivative of:

y = x / sqrt(1-x^2)

Answer 1

quotient rule

Answer 2

Did it but Im messing up somewhere

Answer 3

what did you get?

Answer 4

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

Answer 5

let's try it a different way...
\[\frac{ d }{ dx }\left( \frac{ x }{ \sqrt{1-x^2} } \right)=\frac{ d }{ dx }x\left( \sqrt{1-x^2}\right)^{-1} \]\[= 1\cdot \left( \sqrt{1-x^2}\right)^{-1} + x \cdot (-1)\left( \sqrt{1-x^2}\right)^{-2}\cdot (-2x)\]

Answer 6

Hold on, just working through it

Answer 7

don't be scared of the quotient rule
you probably need it over one denominator in any case, especially if you have to do something with the derivative once you find it

Answer 8

alright so 2x^2 in the numerator then combine the two sqrt(1-x^2) quantities in the denom?

Answer 9

\[=\frac{2x^2+\sqrt{1-x^2}}{1-x^2}\]

Answer 10

sat - I did the quotient rule but I think i messed up when pulling out like terms

Answer 11

don't forget that you need a common denominator

Answer 12

pg can i draw out what i did and you check the work?

Answer 13

shoot, i screwed up because it's a sqrt. should have written it using fractional exponents.

sorry, give me a sec to redo

Answer 14

alright

Answer 15

Use the quotient rule.. I understand it completely just dont know where i went wrong

Answer 16

\[\frac{ d }{ dx }\left( \frac{ x }{ \sqrt{1-x^2} } \right)=\frac{ d }{ dx }x\cdot (1-x^2)^{-\frac{ 1 }{ 2 }}\]\[=(1-x^2)^{-\frac{ 1 }{ 2 }}+x\cdot \left(-\frac{ 1 }{ 2 }\right)(1-x^2)^{-\frac{ 3 }{ 2 }}(-2x)\]