# Download e-book for kindle: A Textbook of Engineering Mathematics-I, 2nd Edition by H.S. Gangwar, Dr. Prabhakar Gupta

By H.S. Gangwar, Dr. Prabhakar Gupta

ISBN-10: 8122424759

ISBN-13: 9788122424751

Written for the scholars of BTech I yr of UP Technical college, Lucknow and different states, this publication discusses intimately the options and strategies in Engineering arithmetic.

**Read Online or Download A Textbook of Engineering Mathematics-I, 2nd Edition PDF**

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

**Sample text**

T. x, we get y1 = cos(a sin–1 x). (v) Replace n by (n – 2) in relation (v), we get yn (0) = {(n – 2)2 – a2}yn–2 (0) Again replace n by (n – 4) in equation (v) and putting yn–2 (0) in above relation, we get yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2}yn–4 (0) Case I. When n is odd: yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... (vi) [The last term in (vi) obtain by putting n = 1 in equation (v)] Putting x = 0, in equation (i), we get y1 (0) = cos(a sin–1 0). a = cos 0 . a ⇒ y1 (0) = a yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} .....

Dz without actually substituting the values of x dt 49 DIFFERENTIAL CALCULUS-I Let δx, δy and δz be the increaments in x, y and z corresponding to a small increament δt in the value of t. + ∂x1 dt ∂x2 dt ∂xn dt The above relation can be also written as In general dz dt = dz = ∂z ∂z dx + ∂y dy , which is called total differential of z. ∂x Corollary: If z = f (x, y) and suppose y is the function of x, then f is a function of one independent variable x. Here y is intermediate variable. Identifying t with x in (iii), we get dz dx = ∂z dy ∂z dx ∂z ∂z dy dz + ∂y ⇒ = + .

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 = .

### A Textbook of Engineering Mathematics-I, 2nd Edition by H.S. Gangwar, Dr. Prabhakar Gupta

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