1. Feb 14, 2008

### JWHooper

I was curious about this one thing: if someone can take 14th derivative of cos(x^3 + 5), then is that person a mathematical genius?

2. Feb 14, 2008

### MaWM

There are things that are difficult because they are difficult to understand, and there are things that are difficult because they require alot of work. That problem you posed is conceptually simple and doesn't even require that much work.

A genius is judged by his capacity for understanding, not his capacity for busy-work. That's what accountants are for. ;)

3. Feb 14, 2008

### John Creighto

There's a trick to solving the above using a Taylor series so that you don't have to differentiate the above expression 14 times right?

4. Feb 14, 2008

### JWHooper

Mathematically speaking, yes. But, I was just saying that what if someone could actually take 14th derivative..? That would be a very challenging calculation, but it was just a thought.

5. Feb 14, 2008

### Dragonfall

I don't know about challenging. You can probably do it in less than 10 minutes, at the MOST.

6. Feb 15, 2008

### Gib Z

Well I wouldn't know about 10 minutes, it gets extremely horribly and disastrously ugly not even half way through.
$$y=-9(x^2(2(\cos (5 + x^3))x-3x^2(\sin (5 + x^3))x^2) + 2(\cos (5 + x^3))x^2x) -6(\sin (5 + x^3) + 3x(\cos (5 + x^3))x^2)$$ is just the 5th one or so. I would give it a good few hours, if I was ever desperate enough to do that by hand.

7. Feb 15, 2008

### PowerIso

It gets difficult and I can't think of a logical reason to actually do it by hand, but the simple answer is no. I believe a genius isn't classified by how well they can follow a formula.

8. Feb 15, 2008

Personally, I don't think that genius...

...has anything to do with following a formula or doing a calculation. It's you're ability justify your conclusions that matters. I'd be more impressed by someone that can actually prove that the derivative of cosx is -sinx than someone that can meticulously apply standard formulas to differentiate some complex expression.

9. Feb 15, 2008

### John Creighto

Lets do it right.

1) Find the Taylor series expansion of cos(y) about the point you are differentiating about

2) Make the substitute y=x^2+5

3) Expand using the binomial series

4) Differentiate each term 15 times.

10. Feb 15, 2008

### HallsofIvy

Staff Emeritus
Why differentiate? After you have expanded powers of x3+ 5, you have the Taylor's series for cos(x3+ 5) and can just read the derivatives off the coefficients.

11. Feb 15, 2008

### John Creighto

That will only work if you want to know the derivative at x=0. Besides, term by term differentiation of a Taylor series is easy.

12. Feb 15, 2008

### JWHooper

Well, I think my Calculus class will get into Taylor (& Maclaurin) series next week or so. I'll look into it soon.

13. Feb 15, 2008

### Dragonfall

-243*x*(19683*cos(x^3+5)*x^27+1194102*sin(x^3+5)*x^24-27862380*cos(x^3+5)*x^21-321080760*sin(x^3+5)*x^18+1955673720*cos(x^3+5)*x^15+6265939680*sin(x^3+5)*x^12-9928638720*cos(x^3+5)*x^9-6774768000*sin(x^3+5)*x^6+1479878400*cos(x^3+5)*x^3+44844800*sin(x^3+5))

Now let's never speak of this again.

14. Feb 15, 2008

### John Creighto

Did you use a computer program?

15. Feb 15, 2008

### PowerIso

We can only hope.

16. Feb 16, 2008

### dodo

The computer is a genius!