In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). Partial derivatives are used in vector calculus and differential geometry.
The partial derivative of a function
f
(
x
,
y
,
…
)
{\displaystyle f(x,y,\dots )}
with respect to the variable
x
{\displaystyle x}
is variously denoted by
f
x
′
,
f
x
,
∂
x
f
,
D
x
f
,
D
1
f
,
∂
∂
x
f
,
or
∂
f
∂
x
.
{\displaystyle f'_{x},f_{x},\partial _{x}f,\ D_{x}f,D_{1}f,{\frac {\partial }{\partial x}}f,{\text{ or }}{\frac {\partial f}{\partial x}}.}
Sometimes, for
z
=
f
(
x
,
y
,
…
)
,
{\displaystyle z=f(x,y,\ldots ),}
the partial derivative of
z
{\displaystyle z}
with respect to
x
{\displaystyle x}
is denoted as
∂
z
∂
x
.
{\displaystyle {\tfrac {\partial z}{\partial x}}.}
Since a partial derivative generally has the same arguments as the original function, its functional dependence is sometimes explicitly signified by the notation, such as in:
f
x
(
x
,
y
,
…
)
,
∂
f
∂
x
(
x
,
y
,
…
)
.
{\displaystyle f_{x}(x,y,\ldots ),{\frac {\partial f}{\partial x}}(x,y,\ldots ).}
The symbol used to denote partial derivatives is ∂. One of the first known uses of this symbol in mathematics is by Marquis de Condorcet from 1770, who used it for partial differences. The modern partial derivative notation was created by Adrien-Marie Legendre (1786) (although he later abandoned it, Carl Gustav Jacob Jacobi reintroduced the symbol in 1841).
\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=xyz^2
People,do we know how to solve this?
I'm looking for the explicite solution z=f(x,y) so far I'm unable to solve this .Thinking of it for a half a day without much of the progress.Even though I haven't tryed all dirty...
Hi,
What does it mean to put a partial derivative in first brackets and put a right subscript to it of another variable?
(\frac {\partial Y} {\partial Y})_T
Thanks.
Molu
I've have met partial derivatives and the \nabla symbol, however, I was asked today what was the geometrical representation and meaning of \nabla \times r and \nabla \cdot r where r was a surface in 3D (i.e. r(x,y,z) = ...).
For the first one, I think that the answer might be:
\left(...
This is annoying me as i have the answer on the tip of my pen, just can't write it down. I'm not 100% sure i understand what the question is asking me to do.
Consider the quantity u = e^{-xy} where (x,y) moves in time t along a path:
x = \cosh{t}, \mbox{ } y = \sinh{t}
Use a method...
Hi,
I'm getting confused over a few points on the derivative of a parametric equation.
Say we the world line of a particle are represented by coordinates x^i . We then parametrize this world line by the parameter t. x^i = f^i(t) .
Now here is where I get confused. The partial...
In determining if a function is exact, here is the question. If V=V(T,P) and PV+RT, show that dV = R/PdT - RT/P2 dP. Is dV an exact differential?
Do I go about by taking the derivative of R/PdT with respect to T, etc? I know this is not a difficult function, but I just want to make sure I'm...
I'm trying to figure out this equation.
{\Psi} = Ae^{-a(bx-ct)^2}
I've expanded this to
{\Psi} = Ae^{-ab^2x^2-abxct-ac^2t^2}
When I try to find the derivative I get this
\left(\newcommand {\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }...
can anyone verify that the equations on the following page, http://nsr.f2o.org/equations.htm are corretly solved. The equations are used to find the uncertainity in the calculation of acceleration in my physics lab. The uncertinty (delta a) would be the sum of all of the four equations, which...
I'm supposed to find (assume all these d's are the partial derivative sign, not d)
d^2z/dxdy, d^2z/dx^2, and d^2z/dy^2
The one I can't do is z^2 + sinx = tany
I set it equal to zero, so z^2 + sinx - tany=0
so dz/dx = - Fx/Fz = sec^2y/2z
dz/dy = - Fy/Fz = -cosx/2z
multiply them...
A couple of quickies on the interpertation of the partial derivative I want to clear up with myself.
If we have a parametric function:
r(u,v)= x(u,v)i + y(u,v)j+z(u,v)k
then the partial derivative W.R.T u or v is regarded as the tangent vector, and we can think of it as the speed, or...
To me a derivative and a partial derivatice is the same thing. You just take it with respect to another vairable ... move some things around and solve...
Can someone give me an example explainin what's happening... The difference between the two. I can solve it and i just absorb it , but...
if you are given f(x,y)=x^2+y^2 and y=cos(t) x=sin(t), then when you differentiate f with respect to t, you use the partial derivatives of f with respect to x and y in the process. When i was taught partial derivatives, i was told that we "keep all but one of the independent variables fixed..."...
Find the second-order partial derivatives of the given function. In each case, show that the mixed partial derivatives f_{xy} and f_{yx} are equal.
Function:
f(x,y)=x^{3}+x^{2}y+x+4
My work (Correct me if I am wrong):
\frac{\partial{f}}{\partial{x}}}=3x^{2}+2xy+1...
Under what conditions can you replace a total differential with a partial?
dx/dy -> partial(dx/dy)
in the context of 2 independant variables and multiple dependant variables.
Thanks
Two homework problems I can't get.
(1) The question is find the first partial derivatives of the function. The problem is that the function in this problem is
f(x, y) = \int_{y}^{x} \cos{t^2} dt
The main obstacle is getting past this function. I can't integrate it and neither can my...
We went over this breifly in class and I'm confused on it. Were doing first order only and this is the problem: z = 3x^2*y^3*e^(5x-3y) + ln(2x^2 + 3y^3) I know your susposed to Fx(x,y) and treat X or Y as a constant, depending, upon how you want to start, but I'm still unclear as to how to...
I need the partial derivatives of:
f(x,y) = ∫xy cos(t2) dt
are they simply:
∂f/∂x = -2xcos(x2)
and
∂f/∂y = 2ycos(y2)
or am I completely lost here?
Let there be a function f[x,y]: RxR->R
Is there any connection between the differentiability
(I am not sure that this is the right English term - I meant f[x,y]= a*dx+b*dy +something of smaller order)
and the equality fxy=fyx, where fxy means the derivative of f[x,y] first by y, and than...