1. ## 2nd Derivative....chain Rule....again

Folks,

My head is melted on this one. I know the second derivative involves a chain rule and a product rule. I never seem to get my head around it. Is there a clever technique to do this with ease? The first derivative is easy as I can see that s and t are 'intermediate' variables. Not so obvious on the second derivative.

u(x,y)=w(s(x,y),t(x,y))

$
\displaystyle \frac {\partial u}{\partial x} = \frac{\partial w}{\partial s} \frac{\partial s}{\partial x} +\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}
$

$
\displaystyle \frac{\partial^{2} u}{\partial x^{2}} =\frac{\partial}{\partial x} [\frac{\partial w}{\partial s} \frac{\partial s}{\partial x} +\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}]

$

Im confused on how to develope it from here

Thanks

2. Originally Posted by bugatti79
Folks,

My head is melted on this one. I know the second derivative involves a chain rule and a product rule. I never seem to get my head around it. Is there a clever technique to do this with ease? The first derivative is easy as I can see that s and t are 'intermediate' variables. Not so obvious on the second derivative.

u(x,y)=w(s(x,y),t(x,y))

$
\displaystyle \frac {\partial u}{\partial x} = \frac{\partial w}{\partial s} \frac{\partial s}{\partial x} +\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}
$

$
\displaystyle \frac{\partial^{2} u}{\partial x^{2}} =\frac{\partial}{\partial x} [\frac{\partial w}{\partial s} \frac{\partial s}{\partial x} +\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}]

$

Im confused on how to develop it from here

Thanks
$\displaystyle \frac{\partial^{2} u}{\partial x^{2}} =\frac{\partial}{\partial x} \left(\frac{\partial w}{\partial s} \frac{\partial s}{\partial x} +\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}\right) = \frac{\partial}{\partial x} \left(\frac{\partial w}{\partial s} \frac{\partial s}{\partial x}\right) +\frac{\partial}{\partial x} \left(\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}\right)$

Take these one at a time. First, apply the product rule to $\frac{\partial}{\partial x}\! \left(\frac{\partial w}{\partial s} \frac{\partial s}{\partial x}\right)$:

$\displaystyle \frac{\partial}{\partial x} \left(\frac{\partial w}{\partial s} \frac{\partial s}{\partial x}\right) = \frac{\partial}{\partial x} \left(\frac{\partial w}{\partial s}\right) \frac{\partial s}{\partial x} + \frac{\partial w}{\partial s} \frac{\partial^2 s}{\partial x^2}.$

Next, to evaluate $\frac{\partial}{\partial x}\! \left(\frac{\partial w}{\partial s}\right)$ you have to use the chain rule (this is the bit that people find tricky):

$\displaystyle \frac{\partial}{\partial x} \left(\frac{\partial w}{\partial s}\right) =
\frac{\partial}{\partial s} \left(\frac{\partial w}{\partial s}\right)\frac{\partial s}{\partial x} + \frac{\partial}{\partial t} \left(\frac{\partial w}{\partial s}\right)\frac{\partial t}{\partial x} = \frac{\partial^2 w}{\partial s^2}\frac{\partial s}{\partial x} + \frac{\partial^2 w}{\partial t \partial s}\frac{\partial t}{\partial x}.$

Feed all that back into your original formula. Then you still have to do a similar series of calculations on the other half of the formula, $\frac{\partial}{\partial x}\!\left(\frac{\partial w}{\partial t} \frac{\partial t}{\partial x}\right).$

3. Originally Posted by Opalg
Take these one at a time. First, apply the product rule to $\frac{\partial}{\partial x}\! \left(\frac{\partial w}{\partial s} \frac{\partial s}{\partial x}\right)$:

Thanks Opalg!

But how is it a product rule?

My understanding of a product rule is used for example to find dy/dx when y = u(x) v(x)

but based on your above expression we have

$\diplaystyle \frac{\partial}{\partial x} \left[w(s)*s(x)) \right]$ which leads me to believe its not a product rule because 'w' is not directly a function of 'x' like u and v are as above,

ie it should be

$\diplaystyle \frac{\partial}{\partial x} \left[w(x)*s(x)) \right]$ before we can use the product rule which is not the case we have?
Thanks

4. Originally Posted by bugatti79
My understanding of a product rule is used for example to find dy/dx when y = u(x) v(x)

but based on your above expression we have

$\diplaystyle \frac{\partial}{\partial x} \left[w(s)*s(x)) \right]$ which leads me to believe its not a product rule because 'w' is not directly a function of 'x' like u and v are as above,
That is correct, as far as it goes. In fact, w is not "directly" a function of x. It is a function of s (and t), which are in turn functions of x and y. So it is indirectly a function of x (and y), and to differentiate it partially with respect to x you have to use the chain rule. The other term in the product, s(x), is directly a function of x, so you can differentiate that with respect to x without any further work.

You should already be familiar with procedures of this sort when dealing with functions of a single variable, if you have ever used the technique of implicit differentiation. For example, if y is a function of x and you want to differentiate an expression like $y^2x^2$, then you use the product rule to get $\frac d{dx}(y^2x^2) = 2y\frac{dy}{dx}*x^2 + y^2*2x.$ In that calculation, you have a product of two functions, a function of y and a function of x. You can differentiate it, using the product rule. But in order to differentiate $y^2$ with respect to x, you have to use the chain rule, differentiating it first with respect to y and then multiplying by $\frac{dy}{dx}.$

The calculation for $\frac{\partial}{\partial x}\bigl(\frac{\partial w}{\partial s}\frac{\partial s}{\partial x}\bigr)$ is just the analogous procedure for functions of two variables.

5. Originally Posted by Opalg
That is correct, as far as it goes. In fact, w is not "directly" a function of x. It is a function of s (and t), which are in turn functions of x and y. So it is indirectly a function of x (and y), and to differentiate it partially with respect to x you have to use the chain rule. The other term in the product, s(x), is directly a function of x, so you can differentiate that with respect to x without any further work.

You should already be familiar with procedures of this sort when dealing with functions of a single variable, if you have ever used the technique of implicit differentiation. For example, if y is a function of x and you want to differentiate an expression like $y^2x^2$, then you use the product rule to get $\frac d{dx}(y^2x^2) = 2y\frac{dy}{dx}*x^2 + y^2*2x.$ In that calculation, you have a product of two functions, a function of y and a function of x. You can differentiate it, using the product rule. But in order to differentiate $y^2$ with respect to x, you have to use the chain rule, differentiating it first with respect to y and then multiplying by $\frac{dy}{dx}.$

The calculation for $\frac{\partial}{\partial x}\bigl(\frac{\partial w}{\partial s}\frac{\partial s}{\partial x}\bigr)$ is just the analogous procedure for functions of two variables.
Your example makes sense and its becoming clearer I think. I believe I was thrown off by the fact that we have

$\displaystyle \frac{\partial}{\partial x} \left[w(s(x))*s(x)\right]$ ie the w throws me off... So even though we have the presence of w it is still a product rule? We then differentiate w with chain rule (in correct sequence of course as youv shown)......