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Math Help - 2nd Derivative....chain Rule....again

  1. #1
    Senior Member bugatti79's Avatar
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    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))

     <br />
\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}<br />

     <br />
\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}]<br /> <br />

    Im confused on how to develope it from here

    Thanks
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by bugatti79 View Post
    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))

     <br />
\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}<br />

     <br />
\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}]<br /> <br />

    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) = <br />
\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).
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Opalg View Post
    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
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  4. #4
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    Quote Originally Posted by bugatti79 View Post
    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.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Opalg View Post
    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)......
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