Page 285 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes Multiply both sides of this equation by y, getting

1 ln x 4x 3/2
y = y
2 x

2 1 ln x 4x 3/2
x
= x e x 1
2 x
(Combine the powers of x .)

x 1 x 2 3/2
= (1/2) x e 1 ln x 4x

Example: Differentiate y = x ln x (sec x) 3x
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
ln x
y = x (sec x) 3x
Apply the natural logarithm to both sides of this equation and use the algebraic properties of
logarithms, getting

ln y = ln x ln x (sec ) 3x
x
= ln x (ln x) + ln (sec x) 3x
= (ln x)(ln x) + 3x ln (sec x)
2
= (ln x) + 3x ln (sec x)
Differentiate both sides of this equation. The left-hand side requires the chain rule since y
represents a function of x . Use the product rule and the chain rule on the right-hand side. Thus,
beginning with
2
ln y = (ln x) + (3x) ln (sec x)
and differentiating, we get

1 1 1
x
x
x
y = 2(ln )x 3x (sec tan ) (3)ln(sec )
y x sec x
(Divide out a factor of sec x.)
2ln x
x
x
= 3 tanx 3ln(sec )
x
(Get a common denominator and combine fractions on the right-hand side.)
2lnx x x
x
x
= 3 tanx 3ln(sec )
x x x
x
2lnx 3x 2 tanx 3 ln(sec )
x
=
x
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