Re: Partial derivative of single variable function?
- From: "[Mr.] Lynn Kurtz" <kurtz@xxxxxxxxxxxxxxx>
- Date: Tue, 16 Dec 2008 11:55:04 -0700
On Tue, 16 Dec 2008 18:54:59 +0100, "mlt" <asdf@xxxxxxx> wrote:
"[Mr.] Lynn Kurtz" <kurtz@xxxxxxxxxxxxxxx> wrote in message
news:X+RHSVtq3mIV92FPUKUgmz3kVcgb@xxxxxxxxxx
On Tue, 16 Dec 2008 15:27:54 +0100, "mlt" <asdf@xxxxxxx> wrote:
If we have the following single variable function:
f(x) = sin(x)
Is it true that the partial derivative of f(x) equals f(x) ' ?
When you say partial derivative, you must specify with respect to what
variable. If you mean partial derivative of f(x) with respect to x,
then yes, it is the same as f'(x). If it is taken with respect to some
other variable, the answer is 0.
par f / par x = f(x) ' = cos(x)
which also means that the gradient of a single variable function also
equals
f'(x):
gradient(f) = par f / par x = f'(x) = cos(x)
Not if your gradient is the vector operator "del". Then:
def(f) = f_x i + f_y j + f_z k = f'(x) i + 0 j + 0 k.
But taken a step further would that not give:
...= f'(x) i + 0 j + 0 k = [f'(x) , 0]
It would give [f'(x), 0, 0]. But that's a 3-D vector, not the scalar
f'(x).
--Lynn
http://math.asu.edu/~kurtz
.
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