202 CHAPTER 7. THE DERIVATIVE

x0 and replace the above algorithm with the simpler one xk+1 = xk−Df (x0)−1 (f (xk)).

Then if T x= x−Df (x0)−1 (f (x)) , it follows that as long as x is sufficiently close to

x0,DT (x) = I−Df (x0)−1 Df (x) and the norm of this transformation is very small so

one can use the mean value inequality to conclude that T is a contraction mapping andprovide a sequence of iterates which converge to a fixed point. Actually, the situation willbe a little more complicated because we will do the implicit function theorem first, but thisis the idea.

7.13 Exercises1. Here are some scalar valued functions of several variables. Determine which of these

functions are o(v). Here v is a vector in Rn, v = (v1, · · · ,vn).

(a) v1v2

(b) v2 sin(v1)

(c) v21 + v2

(d) v2 sin(v1 + v2)

(e) v1 (v1 + v2 + xv3)

(f) (ev1 −1− v1)

(g) (x ·v) |v|

2. Here is a function of two variables. f (x,y) = x2y+ x2. Find D f (x,y) directly fromthe definition. Recall this should be a linear transformation which results from mul-tiplication by a 1×2 matrix. Find this matrix.

3. Let f (x,y) =(

x2 + yy2

). Compute the derivative directly from the definition. This

should be the linear transformation which results from multiplying by a 2×2 matrix.Find this matrix.

4. You have h(x) = g (f (x)) Here x ∈ Rn, f (x) ∈ Rm and g (y) ∈ Rp. where f,gare appropriately differentiable. Thus Dh(x) results from multiplication by a matrix.Using the chain rule, give a formula for the i jth entry of this matrix. How does thisrelate to multiplication of matrices? In other words, you have two matrices whichcorrespond to Dg (f (x)) and Df (x) Call z= g (y) ,y= f (x) . Then

Dg (y) =(

∂z∂y1

· · · ∂z∂ym

),Df (x) =

(∂y∂x1

· · · ∂y∂xn

)Explain the manner in which the i jth entry of Dh(x) is ∑k

∂ zi∂yk

∂yy∂x j

. This is a review

of the way we multiply matrices. what is the ith row of Dg (y) and the jth column ofDf (x)?

5. Find fx, fy, fz, fxy, fyx, fzy for the following. Verify the mixed partial derivatives areequal.

(a) x2y3z4 + sin(xyz)

(b) sin(xyz)+ x2yz

6. Suppose f is a continuous function and f : U → R where U is an open set andsuppose that x ∈U has the property that for all y near x, f (x) ≤ f (y). Prove thatif f has all of its partial derivatives at x, then fxi (x) = 0 for each xi. Hint: Considerf (x+ tv) = h(t). Argue that h′ (0) = 0 and then see what this implies about D f (x).