Why Aren't Partial Derivatives Written with Respect to Composite Variables?

[demonelite123]

Let g(t) = f(tx, ty).

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

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

 

[Simon Bridge]

Because it looks untidy and $t_x = x$ right?

 

[demonelite123]

Because it looks untidy and $t_x = x$ right?

sorry, what do you mean by t_x = x? partial derivative of t with respect to x? could you explain a little more?

 

[Simon Bridge]

sorry, what do you mean by tx=x?

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

g=f(h,k) ... h=tx, k=ty ??

 

[demonelite123]

yes i meant for tx and ty to be t*x and t*y (normal multiplication) respectively.

 

[arildno]

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.

 

[Simon Bridge]

... 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 let's is write, instead: $$g'=x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}$$... does that work?

 

[demonelite123]

... 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 let's is write, instead: $$g'=x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}$$... does that work?

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 f_h and f_k but they are being multiplied by just h and k respectively? shouldn't it be something like f_hh' + f_kk'?

also, thank you for your reply.

 

[Simon Bridge]

for the derivative g' which variable is the derivative with respect to?

look at post #1.

 

[demonelite123]

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: \frac{\partial{f(y + an, y' + an', x)}}{\partial a} = n\frac{\partial f}{\partial y} + n'\frac{\partial f}{\partial y'}.

once again, i am very confused over why didn't instead write: \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'.

further patience with me would be greatly appreciated.