Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.27

Example 58: y = x (x^2 – 1) (x^3 + 2), find dydx.

Let y = uvw, where u = x, v =x^2 – 1, w = x^3 + 2

du dv1, 2x

dx dx= = and

dw 3x 2

dx =

dy ddx dx= (uvw vw uw uv)= du dv dwdx+ dx dx+

= (x^2 – 1) (x^3 + 2).1 + x (x^3 + 2). 2x + x (x^2 – 1). 3x^2
= 6x^5 – 4x^3 + 6x^2 (on simplification).

Example 59: If y = 10x x^10 , find dydx. Let y = u.n where u = 10x and v = x^10

Now du ddx dx= 10 10 log 10 ;x= x e

Again dv ddx dx=. x 10.x 10x.^10 = 10 1− =^9

∴ dy ddx dx= (u.v. u v 10)= dv dudx dx+ = x 10x^9 + x^10. 10x loge 10 = 10x (10x^9 + x^10 loge 10)

REGARDING DIVISION :

Example 60: If y=x 1x 1+− , find dydx

Let y ,=vu where u = x – 1, ddx dxu d= (x 1−)=dx ddx dx− .1 1 0 1= − =

v x 1,= + dx dxdv d= (x 1+)=dx ddx dx+ 1 1 0 1= + =

∴ ( ) ( )

(^2) ( ) (^2) ( ) (^2) ( ) 2
v udu dv
dy dx dx x 1.1 x 1.1 x 1 x 1 2
dx v x 1 x 1 x 1

− + − − + − +

= = = =

+ + +

Example 61: If ( )( )

2

2

x 1 2x 1

y x 1 ,

+ −

= + find dydx.

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

(^) dxdu=6x 4x 1^2 + − and v = x^2 + 1 ; ddxv=2x

( ) ( ) ( )( )

( ) ( )

(^22243)
2 2 2 2 2
v udu dv
dy dx dx x 1 6x 4x 1 x 1 2x 1. 2x 2x 3x 2x 1
dx v x 1 x 1

− + + − − + − + + −

= = =

+ +