Question

Find the derivative dy/dx if

Answer 1

\(sin^{-1}(xy)+\frac{\pi}{2}=cos^{-1}y\)

Answer 2

\(\frac{d}{dx}(sin^{-1}(xy)+\frac{\pi}{2})=\frac{d}{dx}(cos^{-1}(y))\)
\(\frac{1}{\sqrt{1-(xy)^2}}*(1y+xy')=\frac{-1}{\sqrt{1-y^2}}*y'\)
would it be like this?

Answer 3

yes...so far

Answer 4

cool!
\(\frac{y}{\sqrt{1-x^2y^2}}+\frac{xy'}{\sqrt{1-x^2y^2}}=\frac{-1}{\sqrt{1-y^2}}*y'\)
\(\frac{y}{\sqrt{1-x^2y^2}}=\frac{-1}{\sqrt{1-y^2}}*y'-\frac{xy'}{\sqrt{1-x^2y^2}}\)

Answer 5

\(\frac{y}{\sqrt{1-x^2y^2}}=(\frac{-1}{\sqrt{1-y^2}}-\frac{x}{\sqrt{1-x^2y^2}})y'\)
\(\frac{y}{\sqrt{1-x^2y^2}}=(\frac{-\sqrt{1-x^2y^2}}{\sqrt{1-y^2}(\sqrt{1-x^2y^2})}-\frac{x(\sqrt{1-y^2})}{\sqrt{1-x^2y^2}(\sqrt{1-y^2})})y'\)

Answer 6

\(\frac{y}{\sqrt{1-x^2y^2}}=(\frac{-\sqrt{1-x^2y^2}-(x)(\sqrt{1-y^2})}{\sqrt{1-x^2y^2}(\sqrt{1-y^2})})y'\)

Answer 7

\(\frac{y(\sqrt{1-y^2})}{-\sqrt{1-x^2y^2}-(x)(\sqrt{1-y^2})}=y'\)
\(\frac{dy}{dx}=\frac{y(\sqrt{1-y^2})}{-\sqrt{1-x^2y^2}-(x)(\sqrt{1-y^2})}\)

Answer 8

Am I correct? :O