Find Second Derivative y''(x)

[getoutofmylaboratory]

Was wondering if someone could walk me through this. I need to find the second derivative y''(x).

e^(xy) + 2y - 3x = sin y
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[d/(dx)] e^(xy) + 2y - 3x = [d/(dx)] sin y

Ok, so the derivative of sin y is cos y. From here on out, I'm not sure where to go. Derivative of e^x is e^x, not sure how that helps me with e^(xy). What next?

Thanks

 

[lookagain]

take x derivatives on both sides

\(\displaystyle \frac{d}{dx}\left(e^{xy}+2y-3x\right)=\frac{d}{dx}sin[y]\)

\(\displaystyle ye^{xy}\frac{dy}{dx} \ \ \ \ \) <------ It's more complicated than this. Taking he derivative of the first term,
corresponding to this one, necessitates using the product rule when using the chain rule
to take the derivative of the exponent, xy.

\(\displaystyle + \ 2\frac{dy}{dx}-3=cos[y]\frac{dy}{dx}\)

\(\displaystyle e^{xy} \ + \ 2y \ - \ 3x \ = \ sin(y)\)

\(\displaystyle e^{xy}(1y \ + \ xy') \ + \ 2y' \ - \ 3 \ = \ [cos(y)]y' \ \implies\)

\(\displaystyle ye^{xy} \ + \ xy'e^{xy} \ + \ 2y' \ - \ 3 = \ y'cos(y) \ \implies\)

\(\displaystyle y'[xe^{xy} \ + \ 2 \ - \ cos(y)] \ = \ 3 \ - \ ye^{xy} \ \implies\)


\(\displaystyle y' \ = \ \dfrac{3 \ - \ ye^{xy}}{xe^{xy} \ + \ 2 \ - \ cos(y)} \ \ \ \ or \ \ for \ \ one \ \ of \ \ the \ many \ \ other \ \ equivalent \ \ forms, \)


\(\displaystyle y' \ = \ - \dfrac{ye^{xy} \ - \ 3}{xe^{xy} \ - \ cos(y) \ + \ 2} \ \ *\)



But then there is having to work out y'' with the quotient rule and the many instances of the product rule within that.
Then later you must substitute * above everywhere you see y' in the second derivative, y''.

 

[getoutofmylaboratory]

Thanks for the help...I ended up finishing it off like this: