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My professor didn't even explain how to do this. I asked the TA,and they told me to equal the first derivative to $0$ and solve for $x$ and then see if the second derivative is less than $0$? Can someone explain this problem to me?

DeepSea
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Elsa
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First, we do the implicit derivative to simplify our equation. Because maxima/minima occur when $f'(x)=0$, we take the implicit derivative and set it equal to 0. We should end up with a few maxima/minima. To tell whether if it is the maxima or minima, we use the second derivative test to see whether it is a maxima or minima(If $f''(x)<0$ then it is a maxima ).

Teoc
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Implicit differentiation is how one approaches equations that are implicitly defined. When one writes $\frac{d}{dx}x^3=3x^2$, it is equally valid to write $d(x^3)=3x^2dx$, with the $dx$ being in a sense divided over to the other side in the first equation. This allows you to find previously uncomputable derivatives. For example, if one wishes to find $\frac{dy}{dx}$ for $1=xy+x$, one does $0=d(1)=d(xy+x)=d(xy)+d(x)=xdy+ydx+dx$ by application of the usual derivative rules. One the applies algebraic manipulation to get $\frac{dy}{dx}$ on one side of the equation, yielding, in this case, $\frac{dy}{dx}=-\frac{y+1}{x}$.

Does that make sense?

Stella Biderman
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Yep, that's exactly what you gotta do: ' equal the first derivative to 0 and solve for x and then see if the second derivative is less than 0'... Assuming you know implicit differentiation it should be easy - and in case you don't - you do know it, it's basically the chain rule, differentiate everything by x, applying chain rule to y.

Djordje
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