Natural log and derivates

**ur5pointos2slo** · October 16th 2008, 02:45 PM

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

**free_to_fly** · October 16th 2008, 02:59 PM

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:
$ \begin{array}{l} e^y = 1 + yx^2 \\ e^y \frac{{dy}}{{dx}} = 2xy + x^2 \frac{{dy}}{{dx}} \\ \end{array} $

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

**Chris L T521** · October 16th 2008, 03:02 PM

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:
$ \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:
$ \frac{{dy}}{{dx}} = \frac{{2xy + x^2 }}{{e^y }} $

You forgot a $\frac{\,dy}{\,dx}$ !!

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

--Chris

**free_to_fly** · October 16th 2008, 03:03 PM

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

**ur5pointos2slo** · October 16th 2008, 03:15 PM

Originally Posted by Chris L T521

You forgot a $\frac{\,dy}{\,dx}$ !!

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

--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?

**ur5pointos2slo** · October 16th 2008, 03:17 PM

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.

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