1. Natural log and derivates

I am trying to find dy/dx
The problem states: e^y = 1 + (X^2)(y)..My calculus book doesn't really explain how to get the derivative in terms I am making any sense of.

Can I rewrite this as?

y= ln(1+(x^2)(y))

What rule would I need to apply to get the next step? would this be a chain rule problem or is their some special log rule I am missing. Any help would be appreciated. Thanks

2. I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:

Just differentiate everything on both sides with respect to x:
$\displaystyle \begin{array}{l} e^y = 1 + yx^2 \\ e^y \frac{{dy}}{{dx}} = 2xy + x^2 \frac{{dy}}{{dx}} \\ \end{array}$

Upon re-arranging:
$\displaystyle \begin{array}{l} \frac{{dy}}{{dx}}\left( {e^y - x^2 } \right) = 2xy \\ \frac{{dy}}{{dx}} = \frac{{2xy}}{{e^y - x^2 }} \\ \end{array}$

3. Originally Posted by free_to_fly
I'd differentiate it implicitly rather than trying to re-arrange everything...that way it's a lot easier:

Just differentiate everything on both sides with respect to y:
$\displaystyle \begin{array}{l} e^y = 1 + yx^2 \\ e^y \frac{{dy}}{{dx}} = 2xy + x^2{\color{red}\frac{\,dy}{\,dx}} \\ \end{array}$

Upon re-arranging:
$\displaystyle \frac{{dy}}{{dx}} = \frac{{2xy + x^2 }}{{e^y }}$
You forgot a $\displaystyle \frac{\,dy}{\,dx}$!!

The answer should be $\displaystyle \frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}$

--Chris

4. Yep I realized that soon as I posted it...what a silly mistake lol. Forgot it was implicit on the RHS too.

5. Originally Posted by Chris L T521
You forgot a $\displaystyle \frac{\,dy}{\,dx}$!!

The answer should be $\displaystyle \frac{\,dy}{\,dx}=\frac{2xy}{e^y-x^2}$

--Chris

Thank you both. I do have one more question though if you guys don't mind. How does the X^2 become in the denominator?

6. Actually I just figured it out. haha..Im guessing you brought the dy/dx over and factored it out then divided. I don't think before asking questions sorry.