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(2) The derivative of r at the попzеrо real nuтвer a is Dr.(h) = -h/a². Proof. (1) Compute, p(a + h, b+ k) — p(а,b) — ak — bhl _ (а+h)(b+ k) — ab — аk - bhl |(h, k)| |(h, k)|| |hk|| |(h, k)| ´

(2) The derivative of r at the попzеrо real nuтвer a is Dr.(h) = -h/a². Proof. (1) Compute, p(a + h, b+ k) — p(а,b) — ak — bhl _ (а+h)(b+ k) — ab — аk - bhl |(h, k)| |(h, k)|| |hk|| |(h, k)| ´

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Prove part 2 of lemma 4.2.6

Lemma 4.2.6 (Derivatives of the Product and Reciprocal Func- tions). Define the product function, p: R? → R, p(r, y) = ry, and define the reciprocal function :R – {0} → R, r(x) = 1/r. Then (1) The derivative of p at the point (a, b) e R² is Dp(a,b) (h, k) = ak + bh. (2) The derivative of r at the nonzero real number a is Dr.(h) = -h/a². Proof. (1) Compute, |p(a + h, b+ k) – P(a, b) – ak – bh| _ (a + h)(b+ k) – ab – ak – bh| |(h, k)| |(h, k)| |hk| = |(h, k)| For any value of (h, k), |h| <|(h, k)| and |k| < |(h,k)| by the Size Bounds, so |hk| = |h||k| < |(h, k)|². Therefore, |(h, k)[² p(a + h, b + k) – p(a, b) – ak – bh| |(h, k)| lim lim (A,k) (0,0) (h, k)(0,0) |(h, k)| lim (h, k)- (0,0) |(h, k)| = 0. (2) is left as exercise 4.2.4.
Transcribed Image Text:Lemma 4.2.6 (Derivatives of the Product and Reciprocal Func- tions). Define the product function, p: R? → R, p(r, y) = ry, and define the reciprocal function :R – {0} → R, r(x) = 1/r. Then (1) The derivative of p at the point (a, b) e R² is Dp(a,b) (h, k) = ak + bh. (2) The derivative of r at the nonzero real number a is Dr.(h) = -h/a². Proof. (1) Compute, |p(a + h, b+ k) – P(a, b) – ak – bh| _ (a + h)(b+ k) – ab – ak – bh| |(h, k)| |(h, k)| |hk| = |(h, k)| For any value of (h, k), |h| <|(h, k)| and |k| < |(h,k)| by the Size Bounds, so |hk| = |h||k| < |(h, k)|². Therefore, |(h, k)[² p(a + h, b + k) – p(a, b) – ak – bh| |(h, k)| lim lim (A,k) (0,0) (h, k)(0,0) |(h, k)| lim (h, k)- (0,0) |(h, k)| = 0. (2) is left as exercise 4.2.4.
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