3.32 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS
Calculus
Example 69: Given y= 2x 1x 2++ , find dydx.
Solution:
Let y u,= where
( ) ( ) ( ) ( )
( )^2
x 2 2x 1 2x 1 x 2d d
u 2x 1 du, dx dx
x 2 dx x 2
+ + + − + +
= + =
+
( ) ( )
( )^2 ( )^2
du x 2 .2 2x 1.1 3
dx x 2 x 2
= + − + =
+ +
dy 1 1 x 2.
du 2 u 2 2x 1
= = +
+
( )^2 ( )3/2
dy dy du 1 x 2 3. 3
dx du dx 2 2x 1 x 2 2 2x 1. x 2
= = + =
+ + + +
Example 70: If y = log log log x^2 , find dydx
Solution :
Let y = log u where u = log v and v = log x^2 = 2 logx.
2
dy 1 1 (^1).
du u logv= = =log logx
2
du 1 1 (^1).
dv v= =logx =2 logx
dv 2
dx x=
∴ 2 2
dy dy du dv.. (^1). 1 2. 1
dx du dv dx= =log logx 2 logx x=xlogxlog logx
Example 71: Differentiate x^5 w.r.t.x^2
Solution:
Let y = x^5 , z = x^2
dy 5x , 4 dz dx 2x, 2
dx= dx dx= = so that
dx 1.
dz 2x=
Now dy dy dxdz dx dz=. 5x .=^4 2x 21 5= x.^3