Mathematics
BRAINIAC:

Find the derivative dy/dx if

11 months ago
BRAINIAC:

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

11 months ago
BRAINIAC:

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

11 months ago
Zarkon:

yes...so far

11 months ago
BRAINIAC:

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

11 months ago
BRAINIAC:

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

11 months ago
BRAINIAC:

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

11 months ago
BRAINIAC:

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

11 months ago
BRAINIAC:

Am I correct? :O

11 months ago