Advanced Calculus by David V. Widder

By David V. Widder

Vintage textual content leads from straight forward calculus into extra theoretic difficulties. specific process with definitions, theorems, proofs, examples and workouts. subject matters contain partial differentiation, vectors, differential geometry, Stieltjes vital, endless sequence, gamma functionality, Fourier sequence, Laplace remodel, even more. a variety of graded workouts with chosen solutions.

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Now apply Theorem 3, setting By the continuity of g and h, we see that Δx and Δy tend to zero with Δr. We have Now let Δr approach zero and make use of Definition 5 and Definition 6 to obtain Replacing x0, y0 by their values and dropping subscripts, we have equation (3). Equation (4) is obtained in a similar way. The results are easily remembered by putting them in the following form, analogous to the second equation of this section: EXAMPLE D. f(x, y) = xy, f1 = y, f2 = x Thus the rule for differentiation of a product is the same whether the factors are functions of one or of two variables, a fact which is also evident from the definition of a partial derivative.

EXAMPLE A. F(x, y, z) = x2 + y2 + z2 − 6 Equation (1) now defines the two explicit functions Compute at (1, −1, 2). 2 OTHER CASES The equation treated in elementary calculus, can now be handled by the present method. If this equation defines y as a function of x, we can compute its derivative in terms of F. For, remembering that y is a function of x, we have EXAMPLE B. u = f(x, u). Find . This is a special case of equation (2) where F(x, u) = f(x, u) − u. EXAMPLE C. u = f(g(x, u), h(y, u)). Find .

U = f(g(x, y), h(x, y)) It would be a simple matter to deduce the fundamental rules for obtaining the differentials of sums, products, quotients, etc. In fact, such a procedure would produce a slightly simpler technique than the one we have already developed, insofar as it concerns composite functions. We illustrate by Example B above. Here, from the definition of the differential, we have Substituting in equation (2), we have du = df = f1dg + f2dh Observe now the close similarity of this result with the definition in equation (1).

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