# Two methods of implicit differentiation don't correspond

I recently attempted a question on implicit differentiation twice. I differentiated using one method in the first attempt and then another method in the second attempt but they do not correspond when I plug in values for the variables x and y. Please look at the two methods and tell me what I am doing wrong.

In the first attempt I just took the first derivative of each side of the equation with respect to x. In the second attempt, I took the ln of both sides of the equation first and then found the derivatives of either side of the equation.

Errata: The answer for the first attempt is y' = y(y - e^(x/y)) / (y ^ 2 - xe^(x/y)) • I think you accidentally introduced an extra $y$ into the second term near the end of the first attempt. This aside, the results may look different but really be the same, if you consider that you could replace $e^{x/y}$ with the very different-looking expression $x-y$ because they are the same according to the original equation. – MPW May 19 '16 at 22:06
• Oh yes the extra y – rert588 May 19 '16 at 22:08
• $$e^{x/y}=x-y$$ – Simply Beautiful Art May 19 '16 at 22:08
• Yes. That is the question. – rert588 May 19 '16 at 22:11
• I put the y there by mistake when i was writing out the answers neatly but even without that y, the two attempts do not correspond. – rert588 May 19 '16 at 22:14

On the second to last line of your first attempt, you had $y'=\frac{y(y-e^\frac{x}{y}y)}{y^2-e^{\frac{x}{y}}x}$ where actually it should be $y'=\frac{y(y-e^\frac{x}{y})}{y^2-e^{\frac{x}{y}}x}$. And then if you substitute $e^{\frac{x}{y}}$ with $x-y$, you will get $y'=\frac{y(y-(x-y))}{y^2-(x-y)x}=\frac{2y^2-xy}{y^2-x^2+xy}$, which is the same as the result from your second approach.
• I am pretty sure their first attempt is correct. That $e^{\frac x y}y$ term comes from distributing the $e^{\frac x y}$ over the numerator of the derivative of $\frac x y$. EDIT: Oh, wait, I see. They factored out a $y$, but then forgot to remove the $y$ from the $e^{\frac x y}y$ term. You're right. – Noble Mushtak May 19 '16 at 22:24
• The key thing to notice here is that your derivative with the first method and second method look different, but are actually the same because $e^{x/y}=x-y$, so we can substitute that into the derivative using the first method to show that it's equivalent to the derivative using the second method. – Noble Mushtak May 19 '16 at 22:31