Derivative proof by induction

  • Thread starter ben.tien
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  • #1
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Homework Statement

: Let f1,....fn be n functions having derivatives f'1....f'n. Develop a rule for differentiating the product g = f1***fn and prove it by mathematical induction. Show that for those points x, where none of the function values f1(x),....fn(x) are zero, we have g'(x)/g(x) = (f'1(x)/f1(x))+...(f'n(x))/(fn(x))



Homework Equations


product rule: (f1*f2) = (f'1*f2 + f1*f'2)


The Attempt at a Solution

: So I used the associativity property to bunch up the n functions into n/2 functions : (f'1*f2 + f1*f'2)....(f'n-1*fn + fn-1*f1n) and that's where I got stuck.
 

Answers and Replies

  • #2
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Homework Statement

: Let f1,....fn be n functions having derivatives f'1....f'n. Develop a rule for differentiating the product g = f1***fn and prove it by mathematical induction. Show that for those points x, where none of the function values f1(x),....fn(x) are zero, we have g'(x)/g(x) = (f'1(x)/f1(x))+...(f'n(x))/(fn(x))



Homework Equations


product rule: (f1*f2) = (f'1*f2 + f1*f'2)


The Attempt at a Solution

: So I used the associativity property to bunch up the n functions into n/2 functions : (f'1*f2 + f1*f'2)....(f'n-1*fn + fn-1*f1n) and that's where I got stuck.
Show us what you did. For an induction proof, you need to establish a base case, and then assume that the statement is true when n = k. Then you need to show that when the statement for n = k is true, the statement for n = k + 1 must also be true.
 
  • #3
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Okay. I've established that (f1*f2*f3)' = (f'1*f2 + f1*f'2)f3 + f1f2f'3 = f'1f2f3 + f1f'2f3+ f1f2f'3 and etc. for f1*...*fn. When n=k, (f1*...fk)' = f'1....fk + f1f'2....fk +...+ f1...f'k and for n=k+1 [f'1...fk*f(k+1)] +...+ [f1...f'k*f(k+1)] + [f1...fk*f'(k+1)]. However that seemed too easy and I'm sure this is right.
 
  • #4
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Okay I see it now. Thanks.
 

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