Page 281 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes
1/x
Example: Differentiate y = 3x 2 5
Solution:

Because a variable is raised to a variable power in this function, the ordinary rules of
differentiation do not apply ! The function must first be revised before a derivative can be taken.
Begin with
1/x
y = 3x 2 5
Apply the natural logarithm to both sides of this equation getting

ln y = ln 3x 2 5 1/x

= (1/ )ln 3x x 2 5

ln 3x 2 5
=
x
Differentiate both sides of this equation. The left-hand side requires the chain rule since y
represents a function of x. Use the quotient rule and the chain rule on the right-hand side. Thus,
beginning with
ln 3x 2 5
ln y =
x

and differentiating, we get

1
x
x (6 ) ln 3x 2 5 (1)
1 3x 2 5
y =
y x 2
(Get a common denominator and combine fractions in the numerator.)
6x 2 ln 3x 2 5 3x 2 5
3x 2 5 3x 2 5
=
x 2
1

(Dividing by a fraction is the same as multiplying by its reciprocal.)

6x 2 3x 2 5 ln 3x 2 5 1
=
3x 2 5 x 2

6x 2 3x 2 5 ln 3x 2 5
=
x 2 3x 2 5
Multiply both sides of this equation by y, getting

6x 2 3x 2 5 ln 3x 2 5
y = y
x 2 3x 2 5

274 LOVELY PROFESSIONAL UNIVERSITY