find d^2y/dx^2 of y^2=x^2+2x

Question

anonymous · Answer

find y' first

anonymous · Answer

$$2yy'=2x+2$$
$$y'=\frac{2x+2}{2y}=\frac{x+1}{y}$$

anonymous · Answer

then 
$$y''=\frac{y-(x+1)y'}{y^2}$$ and now replace y' by what you know it to be

anonymous · Answer

how did you get $$y \prime \prime $$?

anonymous · Answer

quotient rule

anonymous · Answer

okay i got that, then what?

anonymous · Answer

$$(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$$ with 
$$f(x)=x+1,f'(x)=1,g(x)=y, g'(x)=y'$$

anonymous · Answer

now some annoying algebra. you get 
$$\frac{y-(x+1)\times \frac{x+1}{y}}{y^2}$$ multiply top and bottom by y to clear the compound fraction get 
$$\frac{y^2-(x+1)^2}{y^3}$$

anonymous · Answer

and then you are still not done.

anonymous · Answer

because if you multiply out in the numerator you get 
$$y^2-(x^2+2x+1)=y^2-y^2-1=-1$$

anonymous · Answer

so your "final answer" i think is 
$$y''=-\frac{1}{y^3}$$

anonymous · Answer

how did you get the second $$y ^{2}$$

anonymous · Answer

in $$y ^{2}-y ^{2} -1$$

anonymous · Answer

your original equation was 
$$y^2=x^2+2x$$ so replaced 
$$x^2+2x$$ by 
$$y^2$$

anonymous · Answer

OMG!!! thats some hard shxt

anonymous · Answer

oh okay that makes sense 
thank you so much!

anonymous · Answer

there is another way to do it but it is not really any easier, 
always look to the original equation when you are done, because often it is cooked up to simplify a bunch

anonymous · Answer

yw