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I have been doing that work that requires me to use the chain rule on second order partial derivatives to replace variables (x, y) with (u, v) where u and v are functions of x and y. My question is whether the technique below is allowed and why.Image of techniqueTaken from http://www.ucl.ac.uk/~ucahmdl/LessonPlans/Lesson5.pdf

I didn't think that you'd be allowed to factorise a function of the variable that you will differentiate partially with respect to (can i take out functions of x from partial derivatives with respect to x) as shown in the image?

Any help would be greatly appreciated, thank you.

my question in particular: My question

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  • $\begingroup$ I think it would be helpful if the image included a bit more context on how $u$ and $v$ depend of $x, y$ and about what function you are trying to derive $\endgroup$ – RGS Oct 19 '18 at 23:03
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    $\begingroup$ have added link to source pdf and also direct picture of my question im trying to relate the technique to, particularly in finding (d^2z)/(dy^2) $\endgroup$ – user3486373 Oct 19 '18 at 23:04
  • $\begingroup$ does "can I take out functions of $x$ from partial derivatives with respect to $x$" refer to the first step taken in the solution of the first image? $\endgroup$ – RGS Oct 19 '18 at 23:08
  • $\begingroup$ yes, im just not sure why i would be able to do that $\endgroup$ – user3486373 Oct 19 '18 at 23:08
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If I understood your question correctly, it may help clarifying what is done right in the first step.

Note that the derivative of $z$ w.r.t. to $y$ is a function, i.e.

$$\frac{\partial z}{\partial y}(u, v)$$

makes sense and we can rename it to, say, $f(u,v) = \frac{\partial z}{\partial y}(u,v) = x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v}$.

Now let me derive $f$ w.r.t. to $x$:

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right)$$

Now what was done was, first distribute the derivative w.r.t. $x$:

$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) + \frac{\partial}{\partial x}\left(2\frac{\partial z}{\partial v} \right)$$

Now "taking $x$ out" is really the product rule $(ab)' = a'b + ab'$:

$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) = \frac{\partial}{\partial x}(x^2)\times \frac{\partial z}{\partial u} + x^2 \times \frac{\partial}{\partial x}\frac{\partial z}{\partial u} = 2x\frac{\partial z}{\partial u} + x^2 \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$$

From there we just use the chain rule to find $ \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$.

Is it clearer now?

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  • $\begingroup$ That helps a lot, upon reading it i didnt see the subtle use of the product rule but understand how it comes in now. Thanks a lot for the help. $\endgroup$ – user3486373 Oct 19 '18 at 23:19

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