can u all help me to differentiate this, i try but still did not get it... find dy/dx and d^2y/dx^2question:
y^2=ye^(x^2)+2x

Question

can u all help me to differentiate this, i try but still did not get it... find dy/dx and d^2y/dx^2question: 
y^2=ye^(x^2)+2x

anonymous · Answer

$$2yy'=y'e ^{x^2}+2$$
$$2yy'=y'e ^{x^2}+y2xe ^{x^2}+2$$
$$2yy'-y'e ^{x^2}=y2xe ^{x^2}+2$$
$$y'(2y-e ^{x^2})=y2xe ^{x^2}+2$$
$$y'=\frac{ y2xe ^{x^2}+2 }{ 2y-e ^{x^2} }$$

anonymous · Answer

I'm assuming $$\frac{ dy^2 }{ dx^2 }$$ is second derivative?

anonymous · Answer

thanks,, but to differentiate the answer again i have no idea..

anonymous · Answer

k let me try...it going to be long

anonymous · Answer

ok tq so much

anonymous · Answer

$$y''=\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+2xy(2x)e ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2} }{ (2y-e ^{x^2})^2 }$$
$$\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+4x^2ye ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2}) }{ (2y-e ^{x^2})^2 }$$

anonymous · Answer

phewww this is what I got so far

anonymous · Answer

the answer just like that?

anonymous · Answer

ok tq so much..now i get it