Question

Ok how do I calculate this partial derivative?

\[\frac{\partial }{\partial n}(\frac{d^n}{dx^n}\sin x )\]

Answer 1

This is a challenge to @ParthKohli who says he can take the partial derivative of any function so I'm trying to stump him lol. =P

Answer 2

lol nice :) he can do it im sure
there is an easy way and a hard way too for this

Answer 3

Thank God, it is not for me.. lol!

Answer 4

Hahaha well it's open to anyone who wants to try, I figured this out last semester when I was bored and curious.

Answer 5

haha sweet

Answer 6

You're not supposed to do such deadly questions when you're bored.

Answer 7

okay maybe il just drop an obvious hint : nth derivative of sinx

Answer 8

I want to help someone figure this out, so I'll give the most obscure thing I can think of that's related to this.

We can consider a 2-dimensional coordinate system with a set of basis vectors {sin x, cos x} so that means these vectors are the same as sine and cosine. Can you find a matrix that maps sinx to cos x and cosx to -sinx?\[\left(\begin{matrix}1 \\ 0\end{matrix}\right), \ \left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 9

interesting, below matrix will do it i think :

0 1
-1 0

Answer 10

Slightly off, that'll almost do it, but it should look like this: \[\left[\begin{matrix}0 &-1 \\ 1& 0\end{matrix}\right]\] That'll map the vector (1,0) to (0,1) and (0,1) to (-1,0) right? Ok, so we have our differentiation matrix! Now can we take the derivative of this with respect to the power it has? Does that even make sense? Hehehe.

Answer 11

looking at this Qn is like im listeing to this
https://www.youtube.com/watch?v=lt994KTqFRI
im like wow !

Answer 12

Interesting, this is really just a rotation matrix haha. Also notice we could have made a differentiation matrix for e^x and it would just be the identity matrix. Maybe see what the differentiation matrix is for e^x*sinx is if you're bored... Hehehe... Ok but back to the real problem at hand.

This doesn't take any linear algebra it just has to notice: \[\frac{d}{dx} \sin x = \cos x = \sin(x+\frac{\pi}{2})\]\[\frac{d^2}{dx^2} \sin x = -\sin x = \sin(x+\pi)\]\[\frac{d^n}{dx^n} \sin x =\sin(x+n\frac{\pi}{2})\]
Now our initial question we began with changes to something much easier.

Answer 13

The only real question I have is, why would anyone ever want to know how to take the derivative of a derivative like this? This sort of implies that there's a 1/2 order derivative to sin(x). Interestingly enough, the linear algebra I brought in really comes in handy because if you take the square root of that rotation matrix earlier and multiply it by the vector representing sin(x) you get the same answer. So at least it's consistent, whatever taking half a derivative is. It's kind of like how if you have x^2 it means x multiplied by itself twice and x^(1/2) is x multiplied by itself a half of a time... right? =P