By H.S. Gangwar, Dr. Prabhakar Gupta

Written for the scholars of BTech I 12 months of UP Technical collage, Lucknow and different states, this ebook discusses intimately the innovations and methods in Engineering arithmetic.

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Extra info for A Textbook of Engineering Mathematics-I, 2nd Edition

Example text

2 Now, 2 2 2 from (iv) 2 Hence proved. Example 9. If u = tan–1 ∂ 2u ∂x∂y Sol. 1 − y 1 e1 + x + y j 2 ×x M + 1 x e +y j MN e1 + x + y j − y x . √ e1 + x + y j 1+ x + y + x y 2 2 2 2 = j 2 2 2 2 2 2 2 −1/2 2 2 2 2 2 2 2 2 2y OP PP Q 28 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∂u ∂y or = x 1 + x2 e1 + x je1 + y j e1 + x 2 . to x ∂ 2u ∂x ∂y = LM OP L x OP ∂ M x 1 MMe1 + y j e1 + x + y j PP = e1 + y j ∂x MM e1 + x + y j PP N Q N Q LM e1 + x + y j1 − x 1 e1 + x + y j 2x OP 1 2 M PP e1 + y j MN e1 + x + y j Q e1 + x + y j – x = 1 1 .

Iii) 18 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Again replace n by (n – 4) in (iii) and putting yn –2 (0) in above then, we get yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} yn – 4 (0) Case I. If n is odd yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} .......... (12 – 4) y1 (0) But y1 (0) = 2 sin (2 cos–1 0) = 2 sin (π) = 0 ∴ yn (0) = 0. Case II. If n is even yn (0) = {(n – 2)2 – 4}{(n – 4)2 – 4} .......... {22 – 4} y2 (0) yn (0) = 0 Hence for all values of n, even or odd, yn (0) = 0. Example 5. Find the nth derivative of y = x2 sin x at x = 0.

Y. t. t. x. (B) 2 Thus, from (A) and (B), we find ∂ 2u ∂ 2u = . ∂y∂x ∂x∂y Hence proved. FG y IJ , verify that ∂ f H xK ∂y∂x F yI f = tan G J H xK 2 Example 2. If f = tan–1 Sol. We have = ∂2 f . (iv) 2 Differentiating (iii) partially with respect to x, we get F GH ∂2 f x ∂ = 2 ∂x∂y ∂x x + y 2 = Example 3. If u(x + y) = x Sol. (v) F ∂u ∂u I + y , prove that G − J H ∂x ∂y K F1 − ∂u − ∂u I . GH ∂x ∂y JK 2 =4 x2 + y2 x+y bx + yga2xf − ex + y ja1f = x + 2xy − y b x + yg bx + y g bx + ygb2yg − ex + y ja1f = y + 2xy − x bx + yg bx + yg 2 = ∂u ∂y = ∂u ∂u + ∂x ∂y = b x + yg = 1− = ex 2 2 2 2 2 2 = 2 2 2 2 2 2 2 2 4xy 2 4xy ( x + y) 2 = (x − y) 2 (x + y) 2 j e bx + y g 2 ex − y j 2 bx − yg = bx + yg bx + y g 2 + 2xy − y 2 − y 2 + 2xy − x 2 2 2 ∴ 3 ∂2 f ∂2 f = .

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