## sha0403 2 years ago 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

1. winterfez

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

2. winterfez

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

3. sha0403

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

4. winterfez

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

5. sha0403

ok tq so much

6. winterfez

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

7. winterfez

phewww this is what I got so far

8. sha0403