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If you have a function, implicitly defined, how would understand the following;

Assume $y=y(x)$, find $y'(0)$ and $y''(0)$.

After I implicitly derivate it the first time, I have an expression with $dy/dx$. Do I just fill in zero for $x$, remove the parts that gets canceled out? I'm not sure if the answer is a number or a function?

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If you perform implicit differentiation on $f(x,y)=0$, your expressions for derivatives will involve x and y. What you now have to do is the (most probably complicated) task of finding the value of y corresponding to x=0 from the original implicit equation, and then substitute those values of x and y into your derivative expressions. – J. M. Nov 15 '10 at 16:59
Would I have to solve it for X than? – Algific Nov 15 '10 at 17:32
up vote 2 down vote accepted

If you have a formula for $\frac{dy}{dx}$ and you want to find the value of $y'(0)$, then first you plug in $x=0$ to the implicit formula defining $y$, and evaluate to get the value of $y$. Then you plug in the value of $y$ you found and $x=0$ into the formula for $\frac{dy}{dx}$ to get the value of $y'(0)$. Similarly for $y''(0)$.

For example, suppose you are looking at the circle of radius $1$, $x^2+y^2=1$. Differentiating implicitly, we have: \begin{align*} \frac{d}{dx}(x^2+y^2) &= \frac{d}{dx}1\\ 2x + 2y\frac{dy}{dx} &= 0\\ 2y\frac{dy}{dx} &= -2x\\ \frac{dy}{dx} &= -\frac{x}{y}. \end{align*} To find $y'(0)$, we first plug in $x=0$ into $x^2+y^2=1$ to find the value of $y$ that corresponds to $x=0$; we get $y^2=1$, so $y=\pm 1$. Thus, we have two points, $(0,1)$ and $(0,-1)$. Then we plug in these values into the formula for $\frac{dy}{dx}$, and we get that in both cases we have $y'(0) = -\frac{0}{\pm 1}=0$. So $y'(0) = 0$.

For the second derivative, we again take derivatives: \begin{align*} \frac{dy}{dx} &= -\frac{x}{y}\\ \frac{d}{dx}\left(\frac{dy}{dx}\right) &= -\frac{d}{dx}\left(\frac{x}{y}\right)\\ \frac{d^2y}{dx^2} &= -\left(\frac{y(x)' - x(y')}{y^2}\right)\\ \frac{d^2y}{dx^2} &= \frac{x\frac{dy}{dx} - y}{y^2}\\ \frac{d^2y}{dx^2} &= \frac{x\left(-\frac{x}{y}\right) - y}{y^2}. \end{align*} The last equality, because $\frac{dy}{dx}=-\frac{x}{y}$. So now we plug in $x=0$, $y=\pm 1$ into this equation, and we get that $y''(0) = -\frac{y}{y^2} = -\frac{1}{y} = -\frac{1}{\pm 1}=\mp 1$. That is, $y''(0)=1$ on $(0,-1)$, and $y''(0)=-1$ on $(0,1)$.

Of course, if your implicit curve has a single point on the vertical line $x=0$, then you only have one value of $y$ to contend with, so the process will be simpler (as far as evaluating).

But: Notice that you don't take the derivative of the function you get after evaluating at $x=0$ to get the second derivative! You have to take the entire function and only after differentiating again do you once again plug in $x=0$ and the value of $y$ in order to find $y''(0)$.

Added: If all you want is to find the values of $y''(0)$ and $y'(0)$, you can avoid some complications in the computation of $\frac{d^2y}{dx^2}$ by starting from before you solved for $y'$. In the example above, after we find that when $x=0$ we have $y=\pm 1$ and $y'(0)=0$, we can start from the second line in the computation of $y'$: $$2x + 2y\frac{dy}{dx} = 0,$$ and after simplifying by cancelling the $2$, we can take derivatives again: just be careful with the product rule: \begin{align*} x + 2yy' &= 0\\ \frac{d}{dx}(x+2yy') &= 0\\ 1 + 2(y'y' + yy'') &= 0\\ 2(y')^2 + yy'' &= -1. \end{align*} Now, plugging in $x=0$, $y'(0)=0$, and $y(0)=\pm 1$, we get $\pm 1y''(0)=-1$, hence $y''(0) = \mp 1$; i.e., $y''(0)=1$ when $y=-1$ and $y''(0)=-1$ when $y=1$. Then you don't have to worry about fractions until you start solving.

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Instead of actually substituting dy/dx in that last part, can't I just use the value I got for it instead? Unless it's gonna be a frikin huge expression? – Algific Nov 15 '10 at 19:26
@Algific: if you only want the values and not formulas, yes. In fact, you can avoid even more fractions. I'll edit the body, give me a few minutes. – Arturo Magidin Nov 15 '10 at 19:38

There are lot of resources on the web and i think you can figure out your problem by reading some sources. Here i give some files in which the concept of implicit differentiation has been written nicely.

Regarding your problem suppose assume you want to calculate the derivative of the function $z=f(x,y)$, (say $z=x^{2}+y^{2}$) where $y$ is a function of $x$, then by chain rule you have $$\frac{dz}{dx} = 2x + 2y \cdot \frac{dy}{dx}$$

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@Algific: I hope the above the hints above are useful. Still if you don't get it please let me know. – anonymous Nov 15 '10 at 17:16
I have managed to find the derivative. It was just the other part I wasn't clear on. How would I start of with the method mention in the comment above? – Algific Nov 15 '10 at 17:30

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