1. ## Partial Derivative Problem

I have two functions
r = g (x,y)
s = h (x, y)
so its given that

x = r^3 - s
y = s^3 -r

g(0,0) = h(0,0) = 1

partial derivitives (sorry wrong notation): dg/dx, dh/dy and d^2g/dx^2
All at 0,0

Thanks, any help would be appreciated

2. Originally Posted by viraj
I have two functions
r = g (x,y)
s = h (x, y)
so its given that

x = r^3 - s
y = s^3 -r

g(0,0) = h(0,0) = 1

partial derivitives (sorry wrong notation): dg/dx, dh/dy and d^2g/dx^2
All at 0,0

Thanks, any help would be appreciated
Use the chain rule here..

For $\displaystyle \frac{\partial g}{\partial x}$, we know by the chain rule that $\displaystyle \frac{\partial g}{\partial x}=\frac{\partial g}{\partial r}\cdot\frac{\partial r}{\partial x}+\frac{\partial g}{\partial s}\cdot\frac{\partial s}{\partial x}$

We can figure out that $\displaystyle \frac{\partial g}{\partial r}=1$.

To find $\displaystyle \frac{\partial r}{\partial x}$, use partial implicit differentiation: $\displaystyle x=r^3-s\implies 1=3r^2\frac{\partial r}{\partial x}\implies \frac{\partial r}{\partial x}=\frac{1}{3r^2}$

We also can figure out that $\displaystyle \frac{\partial g}{\partial s}=0$. We can disregard the $\displaystyle \frac{\partial s}{\partial x}$ term, since it will end up disappearing.

Thus, we now see that $\displaystyle \frac{\partial g}{\partial x}=\left(1\right)\cdot\left(\frac{1}{3r^2}\right)= \frac{1}{3r^2}$

Since, $\displaystyle r=g(x,y)$, we see that $\displaystyle \frac{\partial g}{\partial x}=\frac{1}{3(g(x,y))^2}$

Now evaluate it at (0,0). You should end up with $\displaystyle g_x(0,0)=\frac{1}{3(g(0,0))^2}=\frac{1}{3(1)^2}=\c olor{red}\boxed{\frac{1}{3}}$

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Try to find $\displaystyle \frac{\partial h}{\partial y}$ using the same technique as I used above.

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For finding $\displaystyle \frac{\partial^2g}{\partial x^2}$, recall that by the chain rule, we have $\displaystyle \frac{\partial^2g}{\partial x^2}=\frac{\partial g}{\partial r}\cdot\frac{\partial^2r}{\partial x^2}+\frac{\partial^2 g}{\partial r^2}\cdot\left(\frac{\partial r}{\partial x}\right)^2$ (I think I have this correct...if not, feel free to state what it should be)

Follow the same process for this one as well.

Does this make sense?

--Chris

3. ## Chain Rule

I don't think I understand the Chain Rules yet. Will go some reading and find out the exact theorem when it comes to partial derivatives