Is Taking the 14th Derivative of cos(x^3 + 5) a Mark of a Mathematical Genius?

[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?

 

[MaWM]

There are things that are difficult because they are difficult to understand, and there are things that are difficult because they require a lot 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. ;)

 

[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?

 

[JWHooper]

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?

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.

 

[Dragonfall]

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

 

[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.

 

[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.

 

[adrian_durham]

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.

 

[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.

5) Set x equal to the vale your are differentiating about.

 

[HallsofIvy]

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.

 

[John Creighto]

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.

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.

 

[JWHooper]

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

 

[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.

 

[John Creighto]

-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.

Did you use a computer program?

 

[PowerIso]

We can only hope.

 

[dodo]

The computer is a genius!