By H.S. Gangwar, Dr. Prabhakar Gupta

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**Extra resources for A Textbook of Engineering Mathematics: v. 1**

**Sample text**

If u = sin–1 Sol. Given x and from (i), or y ∂u ∂x = ∂u ∂x = ∂u ∂y = ∂u ∂y = Adding (ii) and (iii), we get x Example 5. If f(x, y) = x2 tan–1 Sol. or ∂u ∂u +y = 0. (iii) x + y2 Hence proved. F xI GH y JK ∂f = 2x · tan–1 ∂x FG y IJ H xK ⋅ 2 ∂u ∂u +y = 0. (ii) x + y2 F– x I R| F x I U| GH y JK S|1 − GH y JK V| T W x 2 2 1 − . 2 yx − j FG − y IJ F yI H x K 1+G J H xK 1 + y . (i) then prove that 1 1+ FG y IJ H xK yx 2 x2 + y2 – 2 × FG − y IJ H xK 2 y3 x2 + y2 ∂2 f ∂2 f = . (i) F− x I GH y JK F xI 1+G J H yK 1 2 · 2 25 DIFFERENTIAL CALCULUS-I e x x2 + y2 = x +y 2 2 j – 2y tan–1 F xI GH y JK = x – 2y tan–1 F xI GH y JK .

Anxn+n y –n R| F xI F xI x + a G J + .... + a G J S| y H yK H yK T F xI f (x, y) = x F G J ; degree = n H yK f (x, y) = a y + a y . x + ...... + a y . x R| F x I F x I U| = y Sa + a G J + .... 6 EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS Statement: If f is a homogeneous function of x, y of degree n then x ∂f ∂f + y ∂y = nf. , 2006) Proof. t. (ii) 2 FG y IJ H xK + xn−1 y F′ FG y IJ H xK (from (i)). , xn then x1 ∂f ∂f + x2 ∂x1 ∂x 2 ∂f + ... + xn ∂x = nf. n Corollary 1. If f is a homogeneous function of degree n, then 2 x ∂2 f ∂x 2 ∂2 f ∂2 f 2 + 2xy +y = n (n –1)f.

Y e y x + x2 tan–1 z x+ y + 3 3 y x + . z+x y+ z F xI GH y JK , show that x ∂u ∂ 2u ∂u ∂ 2u 2 + y ∂y = 2u and x + 2xy + ∂x∂y ∂x ∂x2 ∂ 2u = 2u. ∂y2 j sin u = x + y , prove that x F 13 + tan u I GH 12 12 JK . F y2 I ∂ 2u 2 3 . If u = tan G x J , show that x H K ∂x2 22. If F x +y I GH x + y JK . e x+ y 2 1 3 1 3 2 2 tanu ∂ 2u ∂ 2u 2 ∂ u = + 2xy + y 2 2 12 ∂x∂y ∂x ∂y 2 –1 2 ∂f ∂f log x − log y 1 + , show that x + y ∂y + 2f (x, y) = 0. 2 2 ∂x xy x x +y ∂u ∂u 2 5 . If u = sec–1 {(x3 + y3) / (x + y)}, show that x + y ∂y = 2 cot u.