Multivariable Calculus: Directional Derivatives, Differentiablity, Chain Rule

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ChiefKeeper92 said:

Homework Statement


Suppose g : ℝ→ℝ is a twice differentiable function. Define f :R3 → ℝ by
f(x,y,z)=g(x^2 +y^2 +z^2).

a. Show that f is differentiable using an analog to the theorem : If the partial derivatives of x and y exist near (a,b) and are continuous at (a,b) then f is differentiable at (a,b).

b. Let \vec{u} be a unit vector pointing in the direction of the vector -3,0,1. Use the Chain Rule to show that D\vec{u}f(1,2,3) = 0

c. Explain in geometric terms why D\vec{u}f(1,2,3) = 0.

Homework Equations



D\vec{u}f = ∇f \bullet u = abs(∇f) abs(u) cosθ = abs(∇f) cosθ.

∇f=∂f/∂x i + ∂f/∂y j + ∂f/∂z k

dz/dt = (∂f/∂x)*(dx/dt) + (∂f/∂y)*(dy/dt)

Just use the equations you have written down, or explain why you are unable to use them.

RGV
 
ChiefKeeper92 said:
#Error

It is against the PF rules to delete your OP after you have received help. Check your PMs.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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