TUTORIAL 7
DIFFERENTIATION: BASIC CONCEPTS & FORMULAE
Reference: JAC: pp. 193 - 213
1.) Differentiate the following:
a.) y = x 5 + x 6 + 1/x + 1/x 2
b.) y = 5 x 50 + 4x3 + 3 + 8/x – 7/x4
c.) y = 2x 4 + 12x 3 – 4x 2 + 7x – 400
d.) y = 5 x 1/5 – 2x 1/3 + 4/x7
e.) y = 2x 3 + 6x 2 + 49x – 54
f.) 2xy = 3x 2 – 5x + 4
g.) 3y = 4x 4 +2x 2 – 4x + 4/x – 2
h.) x2 y = 3x3 – 2x2 + x -1
2.) Find the first-order and second order derivatives for the following
a.) y = x2 – 3x + 2
b.) y = 4x ½ - 3/x + 7/x2
c.) y = 7/x + 2/x3 – 41
d.) y = 9x 5 – 2x 3 + 24/x – 12/x2
e.) y = ax + b
f.) y = ax2 + bx + c
3.) Differentiate using the Chain Rule: (JAC pp. 229)
a.) y = (3x 2 – 5x + 2) 4
b.) y = 1 / (3x + 7)
c.) y = (1 + x 2 ) ½
d.) y = (3x –5) 4
e.) y = (x 2 + 3x + 5) 3
4.) The Product Rule: (JAC pp. 231)
If y = uv, then dy/dx = u dv/dx + v du/dx
Worked Example:
Differentiate y = x2 (2x + 3) 2
u = x2 and v = (2x + 3) 2 = 4x2 + 12x + 9 = 4x2 + 12 x 1 + 9 x 0
du/dx = (2) x2
-1 = 2x
dv/dx = 2 (4x 2-1) + (1) 12 x 1-1 + (0) 9 x 0-1 = 8x + 12 x0 + 0 = 8x + 12
dy/dx = u dv/dx + v du/dx
= x2
(8x + 12) + (2x + 3) 2 (2x)
= x2 (8x + 12) + 2x (2x + 3) 2
= x2 (4) (2x + 3) + 2x (2x + 3) 2
= (2x + 3) (4x2 + 2x (2x + 3))
= (2x + 3) (4x2 + 4x2 + 6x)
= (2x + 3) (8x2 + 6x)
= (2x + 3) (2x) (4x + 3)
= 2x (2x + 3)(4x + 3)
Differentiate the following using The Product Rule:
a.) y = x (2x + 1) 2
b.) y = (3x + 2)(6x2 – 2)
c.) y = (2x – 3)(2x – 1)
d.) y = (3x
– 8)(5x + 6) 2
e.) y = x2(2x + 1) 3
5.) The Quotient Rule: (JAC pp. 234)
If y = u/v, then dy/dx = ( v du/dx – u dv/dx ) / v2
Worked Example:
Differentiate: y = x/(x+1)
u = x
and v = x + 1 = x1 + x 0
du/dx = 1 (x 1-1 ) = 1 (x 0 ) = 1 (1) = 1;
dv/dx = 1
(x 1-1) + 0 (x 1-0) = 1 + 0 = 1
Using the quotient rule:
dy/dx = ( v du/dx – u dv/dx ) / v2
= [ (x+1) (1) – (x) (1) ]/(x+1) 2
= [ x+1 – x ]/(x+1) 2
= 1/(x+1) 2
Differentiate the following using the Quotient Rule:
a.) y = x 2/(x+4)
b.) y = (4x 2 + 3)/(2x –1)
c.) y = (2x + 1)/(4x – 1)
d.) y = 4x / (x – 1)
e.) y = x/(1+x)
