Question

Help with derivatives

Answer 1

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

Answer 2

I keep messing up and getting the wrong answer!

Answer 3

this looks like a job for quotient rule

Answer 4

so you know your denominator is just squared \[\huge \implies \frac{numerator}{1 - x^2}\]
agree?

Answer 5

yes

Answer 6

\[y = \frac{ x }{ \sqrt{1-x^2} }\]\[y' = \frac{ g'h-gh' }{ h^2 } where \frac{ g(x) }{ h(x) }\]

Answer 7

then for the first part of the numerator... you copy the denominator and take the derivative of the numerator

so..
\[\sqrt{1-x^2} \times d(x) \implies \sqrt{1-x^2} \times 1 \implies \sqrt{1 - x^2}\]
did you get this part?

Answer 8

yes

Answer 9

then for the second part...copy the numerator then diffeerntiate the denominator
\[x \times d(\sqrt{1-x^2})\]
this is chain rule...
\[x \times \frac{1}{2\sqrt{1-x^2}} \times (-2x)\]
is this what you got?

Answer 10

Following

Answer 11

so... now... simplify that last part i wrote

Answer 12

It is the adding the fractions with the square roots that are messing me up

Answer 13

simplify that last part i wrote first...

Answer 14

kk

Answer 15

\[-2x^{2}/2(1-x)^{3/2}\]

Answer 16

how did you get (1-x)^3/2

Answer 17

I forgot the x^2

Answer 18

But isn't the derivative of \[(1-x^{2})^{1/2} == \frac{ 1 }{ 2 }(1-x^{2})^{3/2} * -2x \]
@lgbasallote

Answer 19

...are you mixing this with integration?

Answer 20

I meant -3/2

Answer 21

Ohhh
Crap

Answer 22

I thought it was ^-1/2

Answer 23

SOrry

Answer 24

\[\Large \frac d{dx} (\sqrt x) \implies x^{\frac 12} \implies \frac 12 x^{\frac 12 - \frac 22} \implies \frac 12 x^{-\frac 12}\]

Answer 25

Yes my bad! . okok

Answer 26

so simplify \[x \times \frac 1{2\sqrt{1-x^2}} \times (-2x)\]
again

Answer 27

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

Answer 28

@lgbasallote