Question

Need help finding the derivative of y=x/sqrt(1-x^2)

Answer 1

Do you know what the quotient rule is? You're going to need it for this problem :)

Answer 2

Yes i do. I tried using that as well as changing it to a product rule but cant get the correct answer

Answer 3

Okay, so if we say f(x)=x and g(x)=sqrt(1-x^2), can you tell me what f'(x) and g'(x) is?

Answer 4

f'(x) = 1 and g'(x) = 1/2 (1-x^2) ^(-1/2) (2x) ?? I was taught the chain rule which is what i did with the g'(x). Im just not sure if i did it correctly.

Answer 5

Your derivative of g(x) isn't quite correct. I'll explain it:
\[g(x)=\sqrt{1-x^2}\]\[g(u)=\sqrt{u} , u(x)=1-x^2\]\[g'(u)=\frac{1}{2}u^{-\frac{1}{2}}, u'(x)=-2x\]\[g'(x)=\frac{1}{2}(1-x^2)^{-\frac{1}{2}}*-2x \]\[g'(x)= -x(1-x^2)^{-\frac{1}{2}} \]

Answer 6

You were close, but you missed a minus sign when differentiating -x^2

Answer 7

Now for the product rule: \[\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Is your answer :)

Answer 13

That is like what i got but these are my choices A) 1-2x^2/(1-x^2)^(3/2) B) 1-2x^2/(1-x^2)^(1/2) C) 1/1-x^2 D) 1/sqrt(1-x^2) or E) None of these

Answer 14

\[A) \frac{ 1-2x ^{2} }{ (1-x ^{2})^{\frac{ 3 }{ 2 }}} B) \frac{ 1-2x ^{2} }{ (1-x ^{2})^{\frac{ 1}{ 2 }}} C) \frac{ 1 }{ 1-x ^{2} } D) \frac{ 1 }{ \sqrt{1-x ^{2}} }\]

Answer 15

So would it be none of these or can you simplify the answer or rearrange it to get one of these answers?

Answer 16

I ran it through WolframAlpha and my answer is correct, but it also gave an answer of: \[ \frac{1}{(1-x^2)^{\frac{3}{2}}} \]

Answer 17

So although I'm hesitant to say so, it looks like the answer is E.

Answer 18

Okay thank you so much! That helped a lot!

Answer 19

Any time! Glad you found it helpful.