# Module Two: Differential Calculus(continued) synopsis of results and problems (student copy)

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1 Module Two: Differential Calculus(continued) synopsis of results and problems (student copy) Srikanth K S 1 Syllabus Taylor s and Maclaurin s theorems for function of one variable(statement only)- problems. Evaluation of Indeterminate forms. Partial derivatives Definition and simple problems, Euler s theorem problems, total derivatives, partial differentiation of composite functions, Jacobians-definition and problems. extreme values of functions of two variables. 2 Taylor s and maclaurin s expansions 1. Taylor s theorem 2. Maclaurin series is a special case of taylor series. This occurs when a = L hopital rule 1

2 Exercises: (i) (ii) x 2 x x 2 4 x 3 3x 2 3x + 1 x 1 x 3 + x 2 5x Problems (i) Let a 0, a 1,..., a n are real numbers such that a n n a n 1 n + + a a 0 1 = 0 Then show that the polynomial function f(x) = a n x n + + a 1 x + a 0 has at least one root in (0, 1). (ii) Show that if 0 < a < b then, 1 a b < log( b a ) < 1 b a (iii) Find the 4th degree taylor polynomial approximating 1 1+x 2

3 (iv) Find the maclaurin s series for e x, sin x, cos x, tan x, log(1+x), log(1 x), tan 1 (x), sin h(x), cos h(x) (v) Prove that e ex ( = e 1 + x + x x3 + 5 ) 8 x (vi) Find the expansion of log(1 + x + x 2 ) (vii) Expand (1 + x) x upto 4th power of x. (viii) Expand log(sec x) (ix) (x) (xi) (xii) (1 + x) 1/x e = e x 0 x 2 x 0 ( tan x x ) 1/x 2 (sin x)x x 0 = e 1/3 2x (tan x)tan x π/4 3 Partial differentiation 1. Some surfaces z(x, y) = x 2 + y 2 z(x, y) = 1 (x 2 + y 2 ) 3

4 z(x, y) = x2 y x 4 +y 2 2. Continuity of a function at a point by double its. For example, f(x, y) = xy is continuous at (0, 0). But, g(x) = x 0 y 0 y/x 1 + (y/x) 2 = m 1 + m 2 xy x 2 +y 2 is not as we see that as the it depends on the slope of the line we approach the point (0, 0). The picture below gives us the intuition. 4

5 (a) f(x, y) = x y x+y. Prove that it does not exists as (x, y) (0, 0) 3. A function f : D R is said to be differentiable at (x 0, y 0 ) if there exists (a, b) R 2 such that f(x 0 + h, y 0 + k) f(x + 0, y 0 ) ah bk = 0 (h,k) (0,0) h2 + k 2 The partial derivatives are f x = h 0 f(x+h) f(x) h and f y = k 0 f(x+k) f(x) k. (a) Provided f(x, y) = x 3 y + e xy2, find f x, f y, f xy, f xx, f yy (b) Prove that f(x, y, z) = (x 2 + y 2 + z 2 ) 1/2 satisfies the equation f xx + f yy + f zz = 0 5

6 4. If partial derivatives of f(x, y) are continuous at (x 0, y 0 ), then f is continuous. 5. Existence of partial derivatives (at some point) does not assure that the function is differentiable at the point. Recall that 0 (x, y) = (0, 0) f(x, y) = xy x 2 +y (x, y) (0, 0) 2 is not continuous at (0, 0) although f x (0, 0) and f y (0, 0) exist. (1) 6. Derivative or Total derivative of the function f(x, y) is defined as df = f f dx + x y dy f(x, y) = 0 defines a surface and the RHS of the above equation gives the tangent plane when evaluated at some particular point (x 0, y 0 ). 7. Differentiation of composite function (chain rule for partial derivatives) Let f(x, y) and x(r, s) and y(r, s). We express r and s as functions of x and y. Then we could use the chain rule as follows, f x = f r r x + f s s x Exercises: (a) Let U = z sin(y/x) where x = 3r 2 + 2s, y = 2r 2 3s 2. Find U r and U s. (b) If x = r cos θ, y = r sin θ and V (x, y) be some function, show that V 2 x + V 2 y = v 2 r + v2 θ r 2 8. Euler s theorem about a homogeneous function: Let f(x, y) be a homogeneous function of degree p that is f(ax, ay) = a p f(x, y). Then, xf x + yf y = pf Exercise: If f(x, y) = x 4 y 2 sin 1 (y/x), prove that xf x + yf y = 6F 9. Problems (a) If x x y y z z = c, prove that when x = y = z, we have z xy = 1 x log(ex) 6

7 ( ) (b) If u = tan 1 x 2 +y 2 x+y, then xu x + yu y = sin 2u 2 (c) If u is a homogeneous function of degree n prove that xu xx + yu xy = (n 1)u x and xu xy + yu yy = (n 1)u y and hence x 2 u xx + 2xyu xy + y 2 u yy = n(n 1)u ( (d) If u = sin 1 x+2y+3z ), show that x 8 +y 8 +z 8 xu x + yu y + zu z = 3 tan u (e) If u = log r where r 2 = (x a) 2 + (y b) 2 + (z c) 2, then u xx + u yy + u zz = 1 r 2 (f) If x = r cos θ and y = r sin θ, then prove that and hence prove that θ xx + θ yy = 0 (g) If x = r cos θ and y = r sin θ, prove that x r = r x, x θ = r 2 θ x r xx + r yy = r2 x + r 2 y r (h) If x = 2r s and y = r + 2s, find u yx in terms of derivatives of u with respect to r and s. 4 Jacobian and independence 1. Differentiation in the implicit case When F (x, y, z) is not expressed explicitly, we do the following: 0 = df = F x dx + F y dy + F z dz (2) 7

8 We assume that there exists a function f(x, y) such that z = f(x, y). Then, dz = z z dx + dy (3) x y On comparison of the above two equations, we find that f x = F x F z 2. Definition of jacobian and f y = F y F z. 3. An example 4. Some results about the Jacobian 8

9 5. Problems (a) If u = x 3 y find u t provided x 5 + y = t and x 2 + y 3 = t 2 (b) If u 2 v = 3x + y and u 2v 2 = x 2y, find u x, u y, v x, v y (c) If u = x+y 1 xy and v = tan 1 (x), are u and v functionally related? (d) Let x and y be functions of u and v. Further, u and v be functions of r and s. Prove that (i) (ii) (x, y) (u, v) (x, y) = (u, v) (r, s) (r, s) (x, y) (u, v) (u, v) (x, y) = 1 (e) If u 3 + v 3 = x + y and u 2 + v 2 = x 3 + y 3, then show that (u, v) (x, y) = y2 x 2 2uv(u v) (f) If ax 2 + 2hxy + by 2 and Ax 2 + 2Hxy + By 2 are independent unless a A = b B = c C 9

10 5 Extreme values of functions of two variable 1. Tests 2. Find the maxima, minima and saddles of the functions: xy + 2x 3y 6 (ans. (3,2) is a saddle) and x 3 xy + y 2 3. Discuss the extreme points of sin(x + y) 4. If at some point (x 0, y 0 ) if D(x 0, y 0 ) > 0 and f xx (x 0, y 0 ) > 0, then f yy (x 0, y 0 ) > 0 5. If the perimeter a triangle is constant, then the area is maximum when the triangle is equilateral. 6. If x + y + z = 1, x > 0, y > 0, z > 0, find the extreme values of xy + yz + zx (end of the document) 10

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