Question

Can someone walk me through calculating the derivative with respect to x of arctan (x/sqrt(x-x^2) please?

Answer 1

I stated the question in correctly... \[\tan^{-1} x/\sqrt{1-x ^{2}}\]

Answer 2

\[\tan y = \frac{x}{\sqrt{1-x^2}}\]
find dy/dx using implicit differentiation

Answer 3

\[\frac{ \delta y }{ \delta x }=\frac{ \delta y }{ \delta u }\frac{ \delta u }{ \delta x }\]

Answer 4

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

Answer 5

so that you have \[\frac{ \delta }{ \delta u }(\tan^{-1}(u))\]

Answer 6

I'm working on it. I'm really struggling with this one

Answer 7

it's a hard on indeed, it has both the chain rule and the product rule in it. And i'm a bit sleepy and rusty on this subject.

Answer 8

y can be rewritten as \[x \times(1-x ^{2})^{-1/2}\] is that correct? The square root in the denominator always throws me for a loop...

Answer 9

And then apply the product rule to that?

Answer 10

Well I believe what you have is not equal to y but to tan(y). So:\[\tan(y) = x \times(1-x^2)^{-1/2}\]

Answer 11

yes you can rewrite it like that