Derivative of Product of n Functions by Induction

[ben.tien]

Homework Statement

: Let f₁,...f_n be n functions having derivatives f'₁...f'_n. Develop a rule for differentiating the product g = f₁***f_n and prove it by mathematical induction. Show that for those points x, where none of the function values f₁(x),...f_n(x) are zero, we have g'(x)/g(x) = (f'₁(x)/f₁(x))+...(f'_n(x))/(f_n(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*f_n + f_n-1*f1_n) and that's where I got stuck.

 

[Mark44]

Homework Statement

: Let f₁,...f_n be n functions having derivatives f'₁...f'_n. Develop a rule for differentiating the product g = f₁***f_n and prove it by mathematical induction. Show that for those points x, where none of the function values f₁(x),...f_n(x) are zero, we have g'(x)/g(x) = (f'₁(x)/f₁(x))+...(f'_n(x))/(f_n(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*f_n + f_n-1*f1_n) 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.

 

[ben.tien]

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.

 

[ben.tien]

Okay I see it now. Thanks.