Math Is Fun Forum

YHWH

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Registered: 2022-06-21

Posts: 1

Derivative.

1.Find a function f: R → R that is three times differentiable on R but not four times.
2.Let the functions f and g: R → R be three times differentiable. Calculate (f .g)^(3).

zetafunc

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Registered: 2014-05-21

Posts: 2,436

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Re: Derivative.

Hi YHWH,

Welcome to the forum.

1.Find a function f: R → R that is three times differentiable on R but not four times.

It might be easiest to start with a function that isn't differentiable at a single point (say, x = 0) and work backwards.

Consider a function which returns 1 for positive values of x and -1 for negative values. What function could you differentiate to get that?

2.Let the functions f and g: R → R be three times differentiable. Calculate (f .g)^(3).

This is a demonstration of Leibniz's rule for differentiation. Use the product rule to determine the first derivative -- then differentiate again, and again, and see if you can spot a pattern.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,632

Re: Derivative.

hi zetafunc,

I'm not fully getting what the second part means.

I'm calling the function f4 = -x when x< 0; = 0 when x=0; and = x when x>1.  This V shaped pair of lines has a gradient discontinuity at x =0 so is not differentiable there.

That gives f3 = -(x^2) / 2 when x< 0 etc.  Both the LH limit and the RH limit as x tends to 0 give the gradient as 0, so this function is everywhere differentiable.

Similarly, f2 = -(x^3)/ 6 etc and f1 = -(x^4) / 24 etc.  So these are both fully differentiable.

As I have dis-regarded constants of integration there are plenty more functions that work like this, so I suppose I could have different f and g, for the second part. But then what?

Does (f.g)^3 means f.f.f times g.g.g or differentiate three times.

I feel I'm on the wrong track entirely.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

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Re: Derivative.

Hi Bob,

No I don't think you are on the wrong track -- your f1 function works (it is three times differentiable on

but not four times) because of the cusp at

, and you are right that there are a range of possible solutions we could have just by adding constants (or even polynomial terms that vanish after taking successive derivatives).

I think the two questions are intended to be separate (the condition that f, g are three-times differentiable is probably just there to ensure that f''' and g''' exist rather than bearing any relation to Q1). I am interpreting (f.g)^(3) as meaning the third derivative of (f times g) but happy to be corrected by the thread starter.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,632

Re: Derivative.

hi zetafunc,

Thanks for the reply.  In that case how about:

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

Website

Re: Derivative.

Hi Bob,

Yes, agreed -- the answer will mirror the binomial expansion of (f + g)^3, with the same binomial coefficients, just with number of derivatives rather than powers.