Reverse Chain Rule (Cambridge O Level Additional Maths)

Revision Note

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Reverse Chain Rule

What is the reverse chain rule?

  • The Chain Rule is a way of differentiating two (or more) functions
  • The Reverse Chain Rule (RCR) refers to integrating by inspection
    • Spotting that chain rule would be used in the reverse (differentiating) process

How do I know when to use the reverse chain rule?

  • The reverse chain rule is used when we have the product of a composite function and the derivative of its second function
  • Integration is trickier than differentiation; many of the shortcuts do not work
    • For example, in general integral straight e to the power of straight f left parenthesis x right parenthesis end exponent space straight d x not equal to fraction numerator 1 over denominator straight f apostrophe left parenthesis x right parenthesis end fraction straight e to the power of straight f left parenthesis x right parenthesis end exponent
    • However, this result is true ifspace straight f left parenthesis x right parenthesis is linearspace left parenthesis a x plus b right parenthesis
  • Formally, in function notation, the reverse chain rule is used for integrands of the form 

I equals integral g to the power of apostrophe open parentheses x close parentheses f to the power of apostrophe open parentheses g open parentheses x close parentheses close parentheses space straight d x

    • This does not have to be strictly true, but ‘algebraically’ it should be
  • If the coefficients do not match ‘adjust and compensate’ can be used
    • For example, straight e to the power of x squared end exponent differentiates to 2 x straight e to the power of x squared end exponent with the  chain rule
      • sointegral 2 x straight e to the power of x squared end exponent space straight d x equals straight e to the power of x squared end exponent plus c with the reverse chain rule
    • But to do integral 5 x straight e to the power of x squared end exponent straight d x  we need to:
      • Take out the five: 5 integral x straight e to the power of x squared end exponent straight d x
      • Force a 2 inside (adjust) and divide the outside by a 2 (compensate): 5 over 2 integral 2 x straight e to the power of x squared end exponent space straight d x
      • The bit inside the integral is now a reverse chain rule
      • The answer is 5 over 2 straight e to the power of x squared end exponent plus c
  • A particularly useful instance of the reverse chain rule to recognise is

I equals integral fraction numerator f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space straight d x equals ln space vertical line f left parenthesis x right parenthesis vertical line plus c

    • i.e.  the numerator is (almost) the derivative of the denominator
    • 'adjust and compensate' may need to be used to deal with any coefficients
      • e.g.  I equals integral fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third integral 3 fraction numerator x squared plus 1 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third integral fraction numerator 3 x squared plus 3 over denominator x cubed plus 3 x end fraction space space straight d x equals 1 third ln space vertical line x cubed plus 3 x vertical line plus c

Examiner Tip

  • You can always check your work by differentiating, if you have time

Worked example

A curve has the gradient functionspace f apostrophe left parenthesis x right parenthesis equals 5 x squared sin left parenthesis 2 x cubed right parenthesis.

Given that the curve passes through the pointspace left parenthesis 0 comma space 1 right parenthesis, find an expression forspace straight f left parenthesis x right parenthesis.

Write f(x) as an integral.

straight f open parentheses x close parentheses space equals space integral 5 x squared space sin space open parentheses 2 x cubed close parentheses space straight d x space

Take 5 out of the integral as a factor. 

straight f open parentheses x close parentheses space equals space 5 integral x squared space sin space open parentheses 2 x cubed close parentheses space straight d x space

The main function is sin(...), which would have come from -cos(...). 

Adjust and compensate the coefficients.
2x3 would differentiate to 6xso -cos(2x3) would differentiate to (6x2)sin(2x3)

straight f open parentheses x close parentheses space equals space 5 space cross times open parentheses 1 over 6 close parentheses cross times integral open parentheses 6 x squared close parentheses sin space open parentheses 2 x cubed close parentheses space straight d x space

Integrate.

straight f open parentheses x close parentheses space equals space open parentheses 5 over 6 close parentheses cross times negative cos open parentheses 2 x cubed close parentheses space plus space c

Integrating Composite Functions (ax+b)

What is a composite function?

  • A composite function involves one function being applied after another
  • A composite function may be described as a “function of a function”
  • This Revision Note focuses on one of the functions being linear – i.e. of the formbold space bold italic a bold italic x bold plus bold italic b

How do I integrate linear (ax+b) functions?

  • The reverse chain rule can be used for integrating functions in the form y = (ax + b)n
    • Make sure you are confident using the chain rule to differentiate functions in the form y = (ax + b)n
    • The reverse chain rule works backwards
  • For n = 2 you will most likely expand the brackets and integrate each term separately
  • If n > 2 this becomes time-consuming and if n is not a positive integer we need a different method completely
  • To use the reverse chain rule integral left parenthesis a x plus b right parenthesis to the power of n d x(provided n is not -1)
    • Raise the power of n by 1
    • Divide by this new power
    • Divide this whole function by the coefficient of x
      • integral left parenthesis a x blank plus blank b right parenthesis to the power of n blank straight d x equals fraction numerator open parentheses a x plus b close parentheses to the power of n plus 1 end exponent over denominator n plus 1 end fraction cross times 1 over a plus c
  • You can check your answer by differentiating it
    • You should get the original function when you differentiate your answer
  • Note that this method only works when the function in the brackets is linear (ax + b)
  • The special cases for trigonometric functions and exponential and logarithmic functions are
    •  space integral sin left parenthesis a x plus b right parenthesis space straight d x equals negative 1 over a cos left parenthesis a x plus b right parenthesis plus c
    •  space integral cos left parenthesis a x plus b right parenthesis space straight d x equals 1 over a sin left parenthesis a x plus b right parenthesis plus c
    • space integral straight e to the power of a x plus b end exponent space straight d x equals 1 over a straight e to the power of a x plus b end exponent plus c
    • space integral fraction numerator 1 over denominator a x plus b end fraction space straight d x equals 1 over a ln open vertical bar a x plus b close vertical bar plus c
  • space c, in all cases, is the constant of integration
  • All the above can be deduced using reverse chain rule
    • However, spotting them can make solutions more efficient

Worked example

Find the following integrals

a)      space integral 3 left parenthesis 7 minus 2 x right parenthesis to the power of 5 over 3 end exponent space straight d x

 

 

Name the integral.

I italic space equals integral 3 open parentheses 7 minus 2 x close parentheses to the power of 5 over 3 end exponent straight d x space equals space 3 integral open parentheses 7 minus 2 x close parentheses to the power of 5 over 3 end exponent straight d x space
 

Using the rule 'raise the power by one, divide by the new power and then multiply by the reciprocal of the derivative' integrate the expression.

I italic space equals space 3 open square brackets fraction numerator 1 over denominator 8 over 3 end fraction open parentheses 7 minus 2 x close parentheses to the power of 8 over 3 end exponent space cross times fraction numerator 1 over denominator negative 2 end fraction close square brackets space plus c

Simplify. 

I italic space equals space 3 open square brackets negative fraction numerator 3 over denominator 16 end fraction open parentheses 7 minus 2 x close parentheses to the power of 8 over 3 end exponent space close square brackets space plus c

b)      space integral 1 half cos left parenthesis 3 x minus 2 right parenthesis space straight d x

 

 

Name the integral.

I italic space equals integral 1 half cos open parentheses 3 x minus 2 close parentheses straight d x space equals space 1 half integral cos open parentheses 3 x minus 2 close parentheses straight d x space space
 

Using the rule space integral cos left parenthesis a x plus b right parenthesis space straight d x equals 1 over a sin left parenthesis a x plus b right parenthesis plus c, integrate the expression. 

I italic space equals space 1 half open square brackets 1 third sin open parentheses 3 x minus 2 close parentheses close square brackets space plus c

Simplify. 

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Paul

Author: Paul

Expertise: Maths

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.