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Calculus

Differentiation Using The Product Rule
Where The Chain Rule Is Needed As Well

Applied To Functions Of Type y = xln(kx + c)
where c and k are constants

 

y = x × ln(kx + c)

dy/dx =
FIRST × differential [SECOND] + differential [FIRST] × SECOND
x × differential [ln(kx + c)] + differential [x] × ln(kx + c)

 


EXAMPLE 1

y = x ln(5x + 9)

dy x 1 5    
=   ×
×  + 1 × ln(5x + 9)
dx   5x + 9      

dy   5x      
=   
   +    ln(5x + 9)
dx   5x + 9      

 


EXAMPLE 2

y = x2 ln(5x + 9)

dy x2 1 5    
=   ×
×  + 2x × ln(5x + 9)
dx   5x + 9      

dy   5x2      
=   
   +    2x ln(5x + 9)
dx   5x + 9      

 


EXAMPLE 3

y = x3 ln(5x + 9)

dy x3 1 5    
=   ×
×  + 3x2 × ln(5x + 9)
dx   5x + 9      

dy   5x3      
=   
   +    3x2 ln(5x + 9)
dx   5x + 9      

 

 

 

 
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EXAMPLE 1
y = x ln(5x + 9)

Let u = x
=> u' = 1

Let v = ln(5x +9)
=> v' = 5/(5x + 9)

y = uv

dy/dx = uv' + u'v

dy/dx
= [x][5/(5x + 9)]
     + [1][ln(5x + 9)]
= 5x/(5x + 9) + ln(5x + 9)

 

EXAMPLE 2
y = x2 ln(5x + 9)

Let u = x2
=> u' = 2x

Let v = ln(5x +9)
=> v' = 5/(5x + 9)

y = uv

dy/dx = uv' + u'v

dy/dx
= [x2][5/(5x + 9)]
     + [2x][ln(5x + 9)]
= 5x2/(5x + 9)
     + 2x ln(5x + 9)

 

EXAMPLE 3
y = x3 ln(5x + 9)

Let u = x3
=> u' = 3x2

Let v = ln(5x +9)
=> v' = 5/(5x + 9)

y = uv

dy/dx = uv' + u'v

dy/dx
= [x3][5/(5x + 9)]
     + [3x2][ln(5x + 9)]
= 5x3/(5x + 9)
     + 3x2 ln(5x + 9)

 
Edited by Dr David Cornelius an Independent Private Maths Tutor with over 25 years of experience and The Secretary of The Association of Tutors in the UK for 15 years.
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