# Thread: Calc 3 help... second partial derivatives and differntials z = f(x,y)

1. ## Calc 3 help... second partial derivatives and differntials z = f(x,y)

I am working on an assignment due tomorrow and am just having a hard time with some of this. I think I am on the right track but not sure.

For the first one it wants me to find the four second partial derivatives and observe that the second mixed partial derivaties are equal.

z = ln(x-y) + (x^3)(y^4)

whats throwing me through a loop is taking a deriv of ln(x-y).... the derivative of ln(u) is U' / U

So my first derivs with respect to Y and X would be:

1 - y / (x-y) + (3x^2)(y^4) for diff of z with respect to x

and

x - 1 / (x-y) + (x^3)(4y^3) for diff of z wrt y

if that much is correct I think the rest is but im not sure on that much.

As far as the second goes I'm not sure at all... I will post it in about an hour to an hour and a half. I am in a bind for time right at the moment but wanna use latex and make it proper so I can get some help.

I thank anyone for help in advance.

2. Originally Posted by battleman13
I am working on an assignment due tomorrow and am just having a hard time with some of this. I think I am on the right track but not sure.

For the first one it wants me to find the four second partial derivatives and observe that the second mixed partial derivaties are equal.

z = ln(x-y) + (x^3)(y^4)

whats throwing me through a loop is taking a deriv of ln(x-y).... the derivative of ln(u) is U' / U

So my first derivs with respect to Y and X would be:

1 - y / (x-y) + (3x^2)(y^4) for diff of z with respect to x

and

x - 1 / (x-y) + (x^3)(4y^3) for diff of z wrt y
Well, I would prefer (1- y)/(x- y) but, yes, those are correct.

if that much is correct I think the rest is but im not sure on that much.

As far as the second goes I'm not sure at all... I will post it in about an hour to an hour and a half. I am in a bind for time right at the moment but wanna use latex and make it proper so I can get some help.

I thank anyone for help in advance.

3. Thank you so much for the confirmation: Now on to the rest

Use the given velocity function to find the acceleration function

v(t) = (6 cos(t)) i + (sin(t) + 1) j

I have no idea on this one... I looked over my notes and they are very unclear. I think that to find the acceleartion you just take the derivative of the velocity? I remember that from physics I think... the acceleartion is the slope of the line tangent to a point on the graph of velocity vs time.

For my last and final problem:

Find z = f(x,y) and use the total differential to approximate the quantity

[3 X (1.97)^2 + 2 X 1.97 X 3.02 + 4 X (3.02)^3] - [ 3 x 2^2 + 2 X 2 X 3 + 4 X 3^3]

For this I have the following:

Delta Z = F (X not + Delta X, Y not + Delta y) - F(X not, Y not)

X not = 3
Y not = 2

I've taken these values from the expression to the right of the minus sign

Z = F(x,y) = [ (X) x Y^2 + (Y)(Y)(X) + 4(X)^3]

Delta X = 3.02 - 3 = 0.02 = dx

Delta Y = 1.97 - 2 = -0.03 = dy

if f is differientiable on an open disk centered at (3,2) then Delta z = dz

f(x,y) = X(Y^2)+ (Y^2)X + 4(X^3)]

F(x,y)' with respect to X : y^2 + y^2 +12(X^2)

F(X,Y)' with respect to Y : 2XY +2YX

These functions are defined and continous everywhere

so therefore

delta Z = dz

dz = f(3,2)' dx WRT X - f(3,2)' dy WRT Y

in english... plug in the value 3,2 for XY into the value of the partial derviative of f with respect to X and subtract from it the partial derviative of f with respect to Y (again setting X to 3 and Y to 2)

my math works out to

dz = [ 2^2 + 2^2 + 12(3^2)](0.02) - [2(3)(2) + (2)(3)(2)] (-0.03)

dz = 3.04

Once again thanks for any and all help. This is due tomorrow and any help I can get is appreciated.

Just so everyone else knows I am allowed to receive help with this.

4. Hint: $\displaystyle \mathbf{a}(t) = \frac{d}{dt}[\mathbf{v}(t)]$.