Paper 4: Fundamentals of Business Mathematics & Statistic

3.26 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

dx dxdy d= (3x 5x 110^5 −^3 + )=dxd(3x^5 )−dxd(5x^3 )+dxd.110

3 x 5 x 0,d^5 d^3

= dx − dx + (as 110 is a constant number)

= 3.5x5 – 1 + 5.3x^3 – 1 = 15x^4 + 15x^2.

(iii) Let y 2 x x= − +4 75 8^5 −^8

dx dxdy d=^ − +2 x x4 75 8^5 −^8 =dxd(−^2 )+dx 5d 4^ x^5 −dx 8d 7^ x^8

0 4 d 7 dx^5 x ,^8

= +5 dx 8 dx− (as – 2 is a constant number)

(^4) .5x5 1 (^7) .8x 4x 7x .8 1 4 7
5 8

= + −− − = −

Example 56: If s = ut + 21 ft ,^2 find dsdt when t = 2.

ds ddt dt 2= (^) ut ft+ (^12) =dtd( )ut+dt 2d 1 ft u (^2) = dt 2 dtdt 1 d+ f t 2
(here u, f, 21 are constants & t is a variable, since we are to differentiate w.r.t. t)
u.1 f.2t u .2ft u ft^1211
2 2

= + − = + = +

For t 2, u 2f.= ddts= +

REGARDING PRODUCT :

Example 57: Differentiate (x + 1) (2x^3 -– 21) with respect to x.

Let y = (x + 1) (2x^3 – 21) = u. n where u = x + 1, n = 2x^3 – 21

dx dxdu d= (x 1+)=dx dxdx d+ .1 1 0 1= + =

dx dxdn d= (2x 21 2 .x^3 − )= dx dxd^3 − d( )21 2.3x 0 6x .= 3 1−− =^2

Now dy d dudx dx dx=u ν+ ν =(x 1. 6x 2x 21.1+)^2 +(^3 − )

= 6x^3 + 6x^2 + 2x^3 – 21 = 8x^3 + 6x^2 – 21.