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Section 13.6 : Chain Rule

11. Determine \({f_{u\,u}}\) for the following situation.

\[f = f\left( {x,y} \right)\hspace{0.5in}x = {u^2} + 3v,\,\,\,\,\,\,\,y = uv\]

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Start Solution

These kinds of problems always seem mysterious at first. That is probably because we don’t actually know what the function itself is. This isn’t really a problem. It simply means that the answers can get a little messy as we’ll rarely be able to do much in the way of simplification.

So, the first step here is to get the first derivative and this will require the following chain rule formula.

\[{f_u} = \frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\]

Here is the first derivative,

\[\frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\left[ {2u} \right] + \frac{{\partial f}}{{\partial y}}\left[ v \right] = 2u\frac{{\partial f}}{{\partial x}} + v\frac{{\partial f}}{{\partial y}}\]

Do not get excited about the “unknown” derivatives in our answer here. They will always be present in these kinds of problems.

Show Step 2

Now, much as we did in the notes, let’s do a little rewrite of the answer above as follows,

\[\frac{\partial }{{\partial u}}\left( f \right) = 2u\frac{\partial }{{\partial x}}\left( f \right) + v\frac{\partial }{{\partial y}}\left( f \right)\]

With this rewrite we now have a “formula” for differentiating any function of \(x\) and \(y\) with respect to \(u\) whenever \(x = {u^2} + 3v\) and \(y = uv\). In other words, whenever we have such a function all we need to do is replace the \(f\) in the parenthesis with whatever our function is. We’ll need this eventually.

Show Step 3

Now, let’s get the second derivative. We know that we find the second derivative as follows,

\[{f_{u\,u}} = \frac{{{\partial ^2}f}}{{\partial {u^2}}} = \frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial u}}} \right) = \frac{\partial }{{\partial u}}\left( {2u\frac{{\partial f}}{{\partial x}} + v\frac{{\partial f}}{{\partial y}}} \right)\] Show Step 4

Now, recall that \(\frac{{\partial f}}{{\partial x}}\) and \(\frac{{\partial f}}{{\partial y}}\) are functions of \(x\) and \(y\) which are in turn defined in terms of \(u\) and \(v\) as defined in the problem statement. This means that we’ll need to do the product rule on the first term since it is a product of two functions that both involve \(u\). We won’t need to product rule the second term, in this case, because the first function in that term involves only \(v\)’s.

Here is that work,

\[{f_{u\,u}} = 2\frac{{\partial f}}{{\partial x}} + 2u\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + v\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial y}}} \right)\]

Because the function is defined only in terms of \(x\) and \(y\) we cannot “merge” the \(u\) and \(x\) derivatives in the second term into a “mixed order” second derivative. For the same reason we cannot “merge” the \(u\) and \(y\) derivatives in the third term.

In each of these cases we are being asked to differentiate functions of \(x\) and \(y\) with respect to \(u\) where \(x\) and \(y\) are defined in terms of \(u\) and \(v\).

Show Step 5

Now, recall the “formula” from Step 2,

\[\frac{\partial }{{\partial u}}\left( f \right) = 2u\frac{\partial }{{\partial x}}\left( f \right) + v\frac{\partial }{{\partial y}}\left( f \right)\]

Recall that this tells us how to differentiate any function of \(x\) and \(y\) with respect to \(u\) as long as \(x\) and \(y\) are defined in terms of \(u\) and \(v\) as they are in this problem.

Well luckily enough for us both \(\frac{{\partial f}}{{\partial x}}\) and \(\frac{{\partial f}}{{\partial y}}\) are functions of \(x\) and \(y\) which in turn are defined in terms of \(u\) and \(v\) as we need them to be. This means we can use this “formula” for each of the derivatives in the result from Step 4 as follows,

\[\begin{align*}\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial x}}} \right) & = 2u\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + v\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial x}}} \right) = 2u\frac{{{\partial ^2}f}}{{\partial {x^2}}} + v\frac{{{\partial ^2}f}}{{\partial y\partial x}}\\ \frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial y}}} \right) & = 2u\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial y}}} \right) + v\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial y}}} \right) = 2u\frac{{{\partial ^2}f}}{{\partial x\partial y}} + v\frac{{{\partial ^2}f}}{{\partial {y^2}}}\end{align*}\] Show Step 6

Okay, all we need to do now is put the results from Step 5 into the result from Step 4 and we’ll be done.

\[\begin{align*}{f_{u\,u}} & = 2\frac{{\partial f}}{{\partial x}} + 2u\left[ {2u\frac{{{\partial ^2}f}}{{\partial {x^2}}} + v\frac{{{\partial ^2}f}}{{\partial y\partial x}}} \right] + v\left[ {2u\frac{{{\partial ^2}f}}{{\partial x\partial y}} + v\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right]\\ & = 2\frac{{\partial f}}{{\partial x}} + 4{u^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 2uv\frac{{{\partial ^2}f}}{{\partial y\partial x}} + 2uv\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {v^2}\frac{{{\partial ^2}f}}{{\partial {y^2}}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2\frac{{\partial f}}{{\partial x}} + 4{u^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 4uv\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {v^2}\frac{{{\partial ^2}f}}{{\partial {y^2}}}}}\end{align*}\]

Note that we assumed that the two mixed order partial derivative are equal for this problem and so combined those terms. If you can’t assume this or don’t want to assume this then the second line would be the answer.