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 said: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.
A derivative proof by induction is a mathematical technique used to prove that a certain property holds for all values of a variable in a given range. It involves using the fact that the derivative of a function at a particular point is equal to the limit of the difference quotient as the point approaches that value.
In a regular proof by induction, the property being proved is based on a discrete set of values, while in a derivative proof by induction, the property is based on a continuous range of values. This means that instead of using discrete steps, the proof relies on the use of limits and derivatives.
The steps involved in a derivative proof by induction are:
Some advantages of using a derivative proof by induction include:
No, a derivative proof by induction can only be used to prove properties that are based on continuous values and can be expressed using calculus. It cannot be used to prove properties that are based on discrete values or do not involve calculus.