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Product Rule (Posted on 2005-11-30) Difficulty: 3 of 5
Students in calculus learn the product rule of differentiation as (f*g)'= f'*g+f*g', but a common mistake is taking the product rule as (f*g)'= f'*g'. Usually that answer is wrong, but there are pairs of functions f and g where the wrong product rule generates the right answer.

If f,g is one of those special pairs and f=xn (n≠0), then what is function g?

  Submitted by Brian Smith    
Rating: 4.0000 (1 votes)
Solution: (Hide)
First write a function equation:
f'*g + f*g' = f'*g'
Solving for g:
f'/(f'-f) = g'/g
Integ[f'/(f'-f)]dx = Integ[g'/g]dx
g = e^Integ[f'/(f'-f)]dx

For any function f (except for e^x), we can find its partner using the integral equation above.
Let f=x^n, which make f'=n*x^(n-1), and substitute:
g = e^Integ[n*x^(n-1)/(n*x^(n-1)-x^n)]dx
Simplifying:
g = e^Integ[n/(n-x)]dx
g = e^(-n*ln[n-x])
g = (e^(ln[n-x]))^(-n)
g = (n-x)^(-n)

g = (n-x)^(-n) is the solution, which can be checked by taking its derivative both ways. The derivative is n^2*x^(n-1)*(n-x)^(-n-1).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsPuzzle ThoughtsK Sengupta2022-12-30 08:28:34
SolutionDifferential SolutionOld Original Oskar!2005-11-30 11:55:29
SolutionSolutionBractals2005-11-30 10:16:56
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