Question

if x>1 and y= sin^-1[2x/(1+x^2)], then dy/dx equals?

Answer 1

Do you mean arcsin or the multiplicative inverse of sin(..) ?

Answer 2

\[y = \sin^{-1} [2x/(1+x ^{2})]\]

Answer 3

let v=2x/(1+x^2) then take the derivative and back sub after, less messy

Answer 4

That doesn't clarify it. Do you mean the inverse function of sin or do you mean the reciprocal of its value?

Answer 5

its sin inverse function!!!!!

Answer 6

\[ \frac{dy}{dx}=\left( 2\, \left( 1+{x}^{2} \right) ^{-1}-4\,{\frac {{x}^{2}}{
\left( 1+{x}^{2} \right) ^{2}}} \right) {\frac {1}{\sqrt {1-4\,{
\frac {{x}^{2}}{ \left( 1+{x}^{2} \right) ^{2}}}}}}
\]