22.6. EXERCISES 479

This is an example of the chain rule. Abusing notation slightly, regard f as a functionof position in the plane. This position can be described with any set of coordinates. Thusf (x,y) = f (r,θ) and so

fx = frrx + fθ θ x.

This will be done if you can find rx and θ x. However you must find these in terms of r andθ , not in terms of x and y. Using the chain rule on the two equations for the transformationin 22.14,

1 = rx cosθ − (r sinθ)θ x, 0 = rx sinθ +(r cosθ)θ x

Solving these using Cramer’s rule,

rx = cos(θ) , θ x =−sin(θ)

r

Hence fx in polar coordinates is

fx = fr (r,θ)cos(θ)− fθ (r,θ)(

sin(θ)r

)

22.6.2 The Derivative of the Inverse FunctionExample 22.6.5 Let f : U → V where U and V are open sets in Rnand f is one to oneand onto. Suppose also that f and f−1 are both differentiable. How are Df−1 and Dfrelated?

This can be done as follows. From the assumptions, x = f−1 (f (x)). Let Ix= x.Then by Example 22.2.7 on Page 464 DI = I. By the chain rule,

I = DI = Df−1 (f (x))(Df (x)) , I = DI = Df(f−1 (y)

)Df−1 (y)

Letting y = f (x), the second yields

I = Df (x)Df−1 (f (x)) .

Therefore,Df (x)−1 = Df−1 (f (x)) .

This is equivalent toDf(f−1 (y)

)−1= Df−1 (y)

orDf (x)−1 = Df−1 (y) ,y = f (x) .

This is just like a similar situation for functions of one variable. Remember(f−1)′ ( f (x)) = 1/ f ′ (x) .

Suppose y = f (x) so that x= f−1 (y). Then the using the repeated index convention,the above can be written as

δ i j =∂xi

∂yk(f (x))

∂yk

∂x j(x) .