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Confusion on the chain rule

  1. Oct 11, 2012 #1
    Let g(t) = f(tx, ty).

    Using the chain rule, we get [itex] g'(t) = (\frac{\partial f}{\partial x})(tx, ty) * x + (\frac{\partial f}{\partial y})(tx, ty) * y [/itex]

    this was actually part of a proof and what i don't understand is that why didn't they write [itex] (\frac{\partial f}{\partial (tx)}) [/itex] and [itex] (\frac{\partial f}{\partial (ty)}) [/itex]? i know that the factors of x and y come from [itex] (\frac{\partial (tx)}{\partial t}) [/itex] and [itex] (\frac{\partial (ty)}{\partial t}) [/itex] respectively, but why aren't the other 2 partial derivatives with respect to tx and ty? what happened to the t's?
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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    Because it looks untidy and ##t_x = x## right?
  4. Oct 11, 2012 #3
    sorry, what do you mean by [itex] t_x = x [/itex]? partial derivative of t with respect to x? could you explain a little more?
  5. Oct 11, 2012 #4

    Simon Bridge

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    Given the context it looked like x and y were being parameterized.
    Did you not mean "tx" as tx but as ##t\times x##?

    g=f(h,k) ... h=tx, k=ty ??
  6. Oct 11, 2012 #5
    yes i meant for tx and ty to be t*x and t*y (normal multiplication) respectively.
  7. Oct 11, 2012 #6


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    This confusion is due to sloppy notation in the book.

    Suppose you have a function f(X,Y). Now, set X=X(x,t)=x*t, Y=Y(y,t)=y*t, so that

    g(t)=f(X(x,t),Y(y,t)) (when we "forget" that g(t) really should be regarded as g(x,y,t)!)

    Now, we get:

    g'(t)=f_(X)*X_(t)+f_(Y)*Y_(t), where the subscript is what we differentiate the function with respect to.
  8. Oct 11, 2012 #7

    Simon Bridge

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    ... yeh: the "prime" is usually differentiation wrt to space not time but it is also a common typo.

    if I use the careful notation on ##g=f(h,k) : h=xt, k=yt## (from post #4)$$g' = hf_h+kf_k = xt\frac{\partial f}{\partial h}+yt\frac{\partial f}{\partial k}$$...so the question becomes: why not then write$$g'=xt\frac{\partial f}{\partial (xt)}+yt\frac{\partial f}{\partial (yt)}$$... that what you are asking?
    Well: $$\frac{\partial f}{\partial h}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial h} = \frac{1}{t}\frac{\partial f}{\partial x}$$... is probably tidier.
    It lets is write, instead: $$g'=x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}$$... does that work?
  9. Oct 12, 2012 #8
    for the derivative g' which variable is the derivative with respect to? also, how come in your expansion of the chain rule for g' you have fh and fk but they are being multiplied by just h and k respectively? shouldn't it be something like fhh' + fkk'?

    also, thank you for your reply.
  10. Oct 12, 2012 #9

    Simon Bridge

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    look at post #1.
  11. Oct 16, 2012 #10
    i've been thinking about this and i am still confused on this issue. i recently came across a similar issue that is really making me doubt my competence with the chain rule.

    my book wants to take the partial derivative of f(y + an, y' + an', x) with respect to a. So it simply writes: [tex] \frac{\partial{f(y + an, y' + an', x)}}{\partial a} = n\frac{\partial f}{\partial y} + n'\frac{\partial f}{\partial y'} [/tex].

    once again, i am very confused over why didn't instead write: [tex] \frac{\partial f(y + an, y' + an', x)}{\partial a} = \frac{\partial f}{\partial (y + an)}\frac{\partial (y + an)}{\partial a} + \frac{\partial f}{\partial (y' + an')}\frac{\partial(y' + an')}{\partial a} = \frac{\partial f}{\partial (y + an)}n + \frac{\partial f}{\partial (y' + an')}n' [/tex].

    further patience with me would be greatly appreciated.
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