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MIT18_024s11_ChCnotes

MIT18_024s11_ChCnotes - Derivatives vector functions of...

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Derivatives - of vector functions. Recall that if x - is a point of R" and if f (5) is a scalar function of - x, then the derivative of f (if it exists) is the vector For some purposes, it will be convenient to denote the derivative of f by a - row matrix rather than by a vector. When we do this, we usually denote the derivative by Df rather than ?f. Thus If we use this notation,.the definition of the derivative takes the following form: where ( ) - 0 as - h -> - 0. Here the dot denotes matrix multiplication, so we must write h - as a column matrix in order for the formula to work;
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è his is the formula that will generalize to vector functions i Definition. Let S be a sybset of R ~ . If - k f : S -> R , then - f(5) is called a vector function -- of a vector variable. In scalar form, we can write - f(5) out in the form Said differently, - f consists of "k real-valued functions of n variables." Suppose now that - f is defined in an open ball about the point - a. We say that - f is differentiable at - a if each of the functions fl(x), ...,fk(~) is differentiable at - a ! (in the sense already defined) . Furthermore, we define the derivative of - f at - a to be the matrix That is, Df(=) is the matrix whose 1- Oth row is the derivative th Dfi (=) of the i- coordinate function of - f. Said differently, - the' derivative 'Df (5) - of'. . - f at - - a is. the k by n matrix whose entry in row i and column j is
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it is often called the Jacobian matrix of - f ( 5 ) . Another notation f o r t h i s matrix is the notation With t h i s notation, many of the formulas w e proved f o r a scalar function f (5) hold without change f o r a vector function - f ( 5 ) . W e consider some of them here: Theorem 1. - The function - f (5) - is differentiable - a t - a -- i f and only - i f where - g(&) -> 0 a s h -> 0 . _I - - \ (Here - f, 5, and - E a r e written a s column matrices.) Proof: Both sides of t h i s equation represent column t h matrices. I f w e consider the I- e n t r i e s of these matrices, we have the following equation: NOW - f is differentiable a t - a i f and only i f each function is. And fi is differentiable a t - a i f and only i f fi Ei (&) -> 0 a s - h -> - 0. But Ei (h) - -> 0 a s h - -> - 0, f o r each i, i f and only if - E (h) - -> - 0 a s h - -> - 0.
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Theorem 2. - If 1 - - f ( x ) - is differentiable - a t - a , - then - f - is continuous - a t a. Proof. I f - f is differentiable a t - a , then so i s each function fi. Then i n particular, fi i s continuous a t - a , whence - f is continuous a t - a. - The general chain - rule. Before considering the general chain rule, let us take the chain r u l e w e have already proved and reformulate it i n t e r m s of matrices. A s s u m e t h a t f (5) = f(xl,.. . ,x ) is a scalar function n defined i n an open b a l l about - a , and t h a t - x (t) = (xl (t) , ... , x n ( t ) ) is a parametrized curve passing through - a. L e t - x ( t O ) = - a. I f f (x) - is differentiable a t - a, and i f - x ( t ) is ) differentiable a t to, and we have shown t h a t the composite f ( ~ ( t ) ) is differentiable a t to, and its derivative is given by the equation when t = t O .
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