How To Find Dz/dx Partial Derivatives At The Point 1 3 1 Along X^2 Yz E^ X-z = 4

Derivatives and second order partial derivatives?

This problem is a lot easier than you think. With partial differentiation, you treat the other variables as constants.

f = sin(xyz)
df/dx = yz*cos(xyz)
The yz is treated like a constant.

d2f/dx2 = -(yz)^2*sin(xyz)
and so on.

With regular differentiation you would have to treat y and z as functions of x, but since x, y and z are all independent, they are not functions of x and hence partial derivitives are used.

Calculus Partial Derivative?

For the first part you need to find

You need to find the partial derivative with respect to x,
the partial derivative with respect to y, and the partial derivative with respect to z.


With partial derivatives you treat all variables as constants unless it is the variable that the derivative is respect.

For example if you are trying to find the partial derivative with respect to x, all other variables such as y,z, or anything else is treated as a constant.


I have the following

partial-x = 2xy^3 + 2yz

partial-y = 3x^2y^2 + 2xz - 3z

partial-z = 2xy -3y

You then after evalute each of these at the point (-2,1,2).

For question 2 the defintion of the total differential would
give

dw = (partial-x) dx + (partial-y) dy + (partial-z) dz

Where dw is the total differential.

Again you need to calculate partial derivatives.

Check: I have the following

(partial-x) = -e^ysin(x)

(partial-y) = e^y(cos x)

(partial-z) = 2z

Note: The problem you gave does not have values for dx, dy,
and dz so you would leave those terms in your equation.

Question 3:

You can approach 2 different ways. You transform the original equation expressed in terms of t and take dw/dt directly (usually not an efficient approach). Or you could use the chain rule. The chain rule would involve this equation.

dw/dt = (dx/dt)(partial-w / partial x) + (dy/dt) (partial - w/ partial-y) + (dz/dt) (partial-w/ partial-z)

You can actually check to see if you did the chain rule correctly by plugging the expressions for 't' back into the original equation and calculating dw/dt directly (you should get the same answer).

Use implicit differentiation to find dz/dy and dz/dx (partial derivatives)?

if f(x,y)=z , F(x,y,z) = f(x,y)-z =0
DF = (DF/dx) dx+(DF/dy )dy+ (DF/dz) dz

DF/dx= (DF/dx) +(DF/dy ) (dy/dx) + (DF/dz) (dz/dx) =0
dy/dx=0 because both x and y are independent

SO
(DF/dx) +(DF/dz) (dz/dx) =0 and finally
dz/dx= - (DF/dx)/ (DF/dz)

In the same way
dz/dy = -(DF/dy)/ (DF/Dz)

In your case , F(x,y,z) =yz + xln(y) - z^2=0

DF/dx= Lny
DF/dz= y-2z
dz/dx= -Lny/ (y-2z)

DF/dy = z+x/y = (yz+x)/y
dz/dy=- (yz+x) / (y^2-2zy)

Finding partial derivatives using implicit differentiation?

Differentiate both sides with respect to x (with y held constant):
(ye^(xy) * cos(yz) + e^(xy) * (-sin(yz) * y ∂z/∂x)) - (y ∂z/∂x e^(yz) * sin(xz)
+ e^(yz) * cos(xz) (z + x ∂z/∂x) + ∂z/∂x = 0.

Solve for ∂z/∂x:
∂z/∂x = [ye^(xy) cos(yz) - ze^(yz) cos(xz)]/ [ye^(xy) sin(yz) + ye^(yz) sin(xz) + xe^(yz) cos(xz) - 1].
----------------
Similarly, differentiate both sides with respect to y (with x held constant):
(xe^(xy) * cos(yz) + e^(xy) * (-sin(yz) * (y ∂z/∂y + z)) - (e^(yz) (y ∂z/∂y + z) * sin(xz)
+ e^(yz) * cos(xz) * x ∂z/∂y) + ∂z/∂y = 0.

Solving for ∂z/∂y:
∂z/∂y = [xe^(xy) cos(yz) - z e^(xy) sin(yz) - ze^(yz) sin(xz)] / [y e^(xy) sin(yz) + ye^(yz) sin(xz) + xe^(yz) cos(xz) - 1].

I hope this helps!

Over a certain region of space, the electric potential is V = 5x –3x^2y + 2yz^2. Find the expression for...?

Given:
V = 5*x - 3*x^2*y + 2*y*z^2

Electric field is the negative gradient of the electric potential function. The negative exists, because a positive test charge will be forced to travel to a lower electric potential, by our reasoning for defining electric potential.

E = -▼V

The upside-down delta represents a vector of partial derivatives:
▼ =

Thus:
E = <-dV/dx, -dV/dy, -dV/dz>

Carry out partial derivatives, remembering that other spatial coordinates are constants if not in the Leibniz notation denominator.
Ex = -dV/dx = -(5 - 6*y*x)
Ey = -dV/dy = -(- 3*x^2 + 2*z^2)
Ez = -dV/dz = -(4*y*z)

Results, after you apply the negations:
Ex = 6*y*x - 5
Ey = 3*x^2 - 2*z^2
Ez = -4*y*z

And you can also construct E as a vector field:
E = <6*y*x - 5, 3*x^2 - 2*z^2, -4*y*z>

Grad f and directional derivative?

grad of f is
basically take derivative of f in terms of x, then y, then z
(y+z, x+z, x+y)

directional derivative at point P is
plug in point

(2,2,2) then along vector field u

so then it is

(2,2,2) dot (2,-1,1)/ mag u

so (2,2,2) dot (2, -1, 1)/ sqrt(6)

which is 4/sqrt(6)



the guy below me for got to divide by the magnitude of ||u|| just a heads up. its an important part of the directional derivative