Find the derivative dy/dx if

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

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

6 months agoyes...so far

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

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

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

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

6 months agoAm I correct? :O

6 months ago