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Differentiation and Its Uses in Business Problems 8 Differentiation and its Uses in Business Problems 8 The objectives of this unit is to equip the learners with differentiation and to realize its importance in the field of business. The unit surveys derivative of a function, derivative of a multivariate functions, optimization of lagrangian multipliers and Cobb-Douglas production function etc. Ample examples have been given in the lesson to demonstrate the applications of differentiation in practical business contexts. The recognition of School of Business differentiation in decision making is extremely important in the filed of business. Blank Page Unit-8 Page-164 Bangladesh Open University Lesson-1: Differentiation After studying this lesson, you should be able to: Explain the nature of differentiation; State the nature of the derivative of a function; State some standard formula for differentiation; Apply the formula of differentiation to solve business problems. Introduction Calculus is the most important ramification of mathematics. The present and potential managers of the contemporary world make extensive uses of this mathematical technique for making pregnant decisions. Calculus is inevitably indispensable to measure the degree of changes relating to different managerial issues. Calculus makes it possible for the enthusiastic and ambitious executives to determine the relationship of different variables on sound footings. Calculus in concerned with dynamic situations, such as how fast production levels are increasing, or how rapidly interest is accruing. The term calculus is primarily related to arithmetic or probability concept. Mathematics resolved calculus into two parts - differential calculus and integral calculus. Calculus mainly deals with the rate of changes in a dependent variable with respect to the corresponding Differential calculus change in independent variables. Differential calculus is concerned with is concerned with the average rate of the average rate of changes, whereas Integral calculus, by its very nature, changes. considers the total rate of changes in variables. Differentiation Differentiation is one of the most important operations in calculus. Its theory solely depends on the concepts of limit and continuity of functions. This operation assumes a small change in the value of The techniques of dependent variable for small change in the value of independent differentiation of a variable. In fact, the techniques of differentiation of a function deal with function deal with the the rate at which the dependent variable changes with respect to the rate at which the independent variable. This rate of change is measured by a quantity dependent variable changes with respect known as derivative or differential co-efficient of the function. to the independent Differentiation is the process of finding out the derivatives of a variable. continuous function i.e., it is the process of finding the differential co- efficient of a function. Derivative of a Function The derivative of a function is its instantaneous rate of change. Derivative is the small changes in the dependent variable with respect to a very small change in independent variable. dy Let y = f (x) , derivative i.e. means rate of change in variable y with dx respect to change in variable x. Business Mathematics Page-165 School of Business The derivative has many applications, and is extremely useful in optimization- that is, in making quantities as large (for example profit) or as small (for example, average cost) as possible. Some Standard Formula for Differentiation Following are the some standard formula of derivatives by means of which we can easily find the derivatives of algebraic, logarithmic and exponential functions. These are : dc 1. = 0, where C is a constant. dx dx n d 2. = [xn] = n. xn-1 dx dx d d 3. a. f() x = a []f() x dx dx 1 −(n +1) d − 1 4. x n = − .x n dx n de x d 5. = e x = e x dx dx () d d 6. [e g() x ]= e g() x . []g() x dx dx dy dy du 7. If y = f(u) and U = g(x) then = × dx du dx d 8. a x = a x .log a dx ( ) e d[ f( x) ± g( x)] d[ f( x)] d[ g( x)] 9. = ± dx dx dx d d 1 10. ()log e x = ln x = dx dx () x n dy n d[ f( x)] 11. If Y = [f(x)] then, = n [f(x)] –1 . dx dx d 1 12. log a x = log e dx x a d[ f( x).g( x)] d[ g( x)] d[ f( x)] 13. = f() x + g() x dx dx dx Unit-8 Page-166 Bangladesh Open University f( x) d d[] f() x − d[] g() x g() x g() x f() x = dx dx 14. 2 dx []g() x d d 15. a g()() x = a g x . []g() x .log e dx dx a du f( x + dx.y) − f( x, y) 16. If U = f(x, y), = and dx dx du f( x, y + dy) − f( x, y) = dy dy de ax 17. If y = e ax, then its first derivative is equal to = e ax dx d 2e as Second derivative is equal to = a2eax dx 2 d 3e ax Third derivative is equal to = a3eax and the nth derivative is dx3 denoted by d n e ax = aneax dx n Derivative of Trigonometrically Functions d d − 18. dx (Sin x) = Cos x. dx (Cos x) = Sinx d 2 d − 2 19. dx (tan x) = Sec x. dx (Cot x) = Cosec x d d − 20. dx (Sec x) = Sec x. tan x; dx (Cosec x) = Cosec x. Cot x d 1 d 1 21. (Sin −1x)= ; (Cos −1x)= − dx 1 − x 2 dx 1 − x 2 d 1 d 1 22. (tan −1x)= ; (Cot −1x)= − dx 1 + x 2 dx 1 + x 2 d 1 d 1 23. (Sec−1x) = ; (Cosec−1x) = − dx x. x 2 −1 dx x x 2 −1 2 2 Sin x Sin x + Cos x = 1; tan x = Cos x Business Mathematics Page-167 School of Business 2 2 Cos x Sec x – tan x = 1; Cot x = Sin x When x and y are separately expressed as the functions of a third variable in the equation of a curve is known as parameter. In such cases dy we can find dx without first eliminating the parameter as follows: Thus, if x =Q (t), y = ψ (t) dy dy dx Then, dt = dx . dt dy dy ∴ = dt dx dx dt Let us illustrate these different derivatives by the following examples. Example – 1: dx If y = f(x) = a; find dy Solution: dy d(a) dx = dx =0, since a is a constant, i.e.,'a' has got no relationship with variable x. Example – 2: Differentiate the following functions, with respect to x, − (i) y = x , (ii) y = 8x 5 (iii) y= 3x 3 − 6x 2+2x − 8 Solution: 1 2 We know that x = x 1 1 −1 dy d −1 1 1 dy 1 x 2 1 2 x 2 dx = dx = .x = = . Hence dx = 2 2 2 x 2 x dy d −5 (ii) dx = dx (8x ) d − 40 = 8 (x−5) = 8( −5) x −6 = − 40x –6 = . dx x6 Unit-8 Page-168 Bangladesh Open University dy − 40 Therefore, = dx x6 dy d 3 − 2 − (iii) dx = dx (3x 6x +2x 8) d d d d 3 − 2 − = dx (3x ) dx (6x ) + dx (2x) dx (8) = 3.3x 3−1− 2.6.x 2−1+ 2 −0 = 9x 2−12x+2 dy 2 − Thus, dx = 9x 12x + 2. Example –3: Differentiate e x(log x) . (2x 2+3) with respect to x. Solution: Let y = e x.(log x) . (2x 2 + 3) dy x d 2 x 2 d 2 d x dx =e (log x) dx (2x +3) +e (2x +3). dx (logx) + (log x). (2x +3). dx (e ) x x 2 1 2 x =e (logx)(4x)+e (2x +3) x + logx(2x +3)e 2 x 2x +3 2 = e [4x.logx + x + (2x +3) logx] 2 dy x 2x +3 2 So, dx = e [4x.logx + x + (2x +3) logx] Example–4: 2+3log x dy If y = , find x2+5 dx Solution: 2+3log x y = x2+5 d d (x 2 + )5 2( + 3log x) − 2( + 3log x) (x 2 + )5 dy dx dx = dx (x 2 + )5 2 3 15 15 (x2+5) ( )–(2+3logx)(2x) –x –6xlogx –x –6xlogx x x dy x = = .Thus, = (x2+5)2 (x2+5)2 dx (x2+5)2 Business Mathematics Page-169 School of Business Example–5: 2 ++ Find the differential co-efficient of e 7x5x with respect to x. Solution: Let y= e x2+5x+y 2 2 dy e ++ yx5x d 2 x + 5x + y. dx = . dx (x +5x +y) = e (2x + 5) 2 dy x + 5x + y dx = (2x + 5). e Example–6: dy Find dx , if y = log 4x+3 Solution: Let y = log 4x+3 1 1 + 2 = log (4x 3) = 2 log (4x +3) dy d 1 1 1 d 1 1 2 dx = dx [2 log (4x +3)] = 2 . 4x+3 . dx (4x +3) = 2 . 4x+3 . 4 = 4x+3 dy 2 ∴ dx = 4x+3 Example–7: 2 Find the first, second and third derivatives when y = x.e x Solution: y = x.e x2 First derivative, dy d x2 x2 d x2 x2 x2 2 dx = x dx e +e . dx (x) = x.e . 2x + e .1 = e (2x +1) d2y Second derivative, = e x2.2x(2x 2+1) +e x2.4x dx 2 2 x 2 2 2 = e x . 4x 3 + e .2x + e x . 4x = e x . (4x 3 + 2x + 4x) = e x2. (4x 3 + 6x) Unit-8 Page-170 Bangladesh Open University Third derivative, d3y = ex2 2x (4x 3 + 6x) + e x2 (12x 2 + 6) dx 3 2 x 2 2 2 = e x .
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