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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?

See The Solution Submitted by Brian Smith    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 2

Let g(x) = C/(n-x)^n.
We want to show that
  [x^n]'*g'(x) = x^n*g'(x) + [x^n]'*g(x)
  ([x^n]' - x^n)*g'(x) = [x^n]'*g(x)
  ([x^n]' - x^n)*g'(x) = (n*x^(n-1) - x^n)*[n*C/(n-x)^(n+1)]
                       = n*x^(n-1)*C/(n-x)^n
                       = [x^n]'*g(x)

  Posted by Bractals on 2005-11-30 10:16:56
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