I have $$\sqrt(1+x^4y^2) = xy+x^2$$ Right now I've got $$(1+x^4y^2)^{-1/2}4x^3yy'=y'+2x$$
The answer is a big fraction but that doesn't seem to be what I' heading towards, I don't know where to go from what I've got, and I'm not sure it's even right. Can some one get me on the right track here?
Given $$(1+x^{4}y^2)^{\frac{1}{2}} = xy+x^2\;,$$ Now Differentiate both side w r. to $x\;,$
$$\displaystyle \frac{d}{dx}\left[(1+x^4y^2)^{\frac{1}{2}}\right] = \frac{x}{dx}(xy+x^2)$$
$$\displaystyle \frac{1}{2}(1+x^4y^2)^{-\frac{1}{2}}\cdot \left[x^4\cdot 2yy'+y^2\cdot 4x^3\right] = xy'+y+2x$$
So $$\displaystyle \left[\frac{x^4y}{\sqrt{1+x^4y^2}}-x\right]y' = y+2x-\frac{2x^3y^2}{\sqrt{1+x^4y^2}}$$
So we get $$\displaystyle y'=\left[\frac{(y+2x)\sqrt{1+x^4y^2}-2x^3y^2}{x^4y-x\sqrt{1+x^4y^2}}\right] = \frac{(y+2x)(xy+x^2)-2x^3y^2}{x^4y-x(xy+x^2)}$$
From first line $(1+x^{4}y^2)^{\frac{1}{2}} = xy+x^2$
Now after simplifivation we get $$\displaystyle y'=\frac{(y+2x)(y+x)-2x^2y^2}{x^3y-xy-x}$$