partial derivative of m sin(2mx+2my)

Question

ksaimouli · Answer

last part is +2ny

ipwnbunnies · Answer

This is easier than you think then.
I was trying to find the partial derivative symbol, but I don't know where it is lol.

ksaimouli · Answer

instead use d i will understand

ipwnbunnies · Answer

So since we're trying to find partial d wrt m, we treat x, m, and y as constants.
Then, it becomes a simple calculus 1 problem.

ipwnbunnies · Answer

dm/d[f(x,y,m,n)] = 2x cos (2mx + 2ny)

ipwnbunnies · Answer

It's not sigma, it's its own special character, and I'm not sure why lol.
Can someone check my derivative?

ksaimouli · Answer

$$m \sin(2mx+2ny)$$

ipwnbunnies · Answer

No, it'll be
$$2xsin(2mx+2ny)$$

ksaimouli · Answer

thats the question

anonymous · Answer

$$
\frac{\partial }{\partial m} \sin(2mx+2my) = \left[\frac{\partial }{\partial (2mx+2my)} \sin(2mx+2my) \right]\left[  \frac{\partial }{\partial m}2mx+2my\right]
$$

ipwnbunnies · Answer

We're treating x, n, and y's as constants in partial derivatives. So, we'll have to use a u-sub on the inside of the argument. 
u = 2mx + 2ny
du/dm = 2x + 0. The term '2ny' is treated as a constant, and we use the Power Rule on the 2xm.

anonymous · Answer

First step is chain rule.

ipwnbunnies · Answer

Hmm, I'm not even good at explaining partial derivaives. I'll let wio take over lol.

ksaimouli · Answer

I used the chain rule$$f_{x}=(m) \cos(2mx+2ny)(2m)$$

anonymous · Answer

We are doing partial derivative with respect to $x$?

ksaimouli · Answer

yes

ipwnbunnies · Answer

Oh I thought you said m.

ipwnbunnies · Answer

Your original post says 'm'.

anonymous · Answer

Your answer looks ok then.

ksaimouli · Answer

this actually fxx (i already found fx) that was the part of the question

ipwnbunnies · Answer

Right. 
$$f_{x} = 2m*\cos(2xm + 2ny)$$

ksaimouli · Answer

okay let me give the question $$(\sin(mx+ny))^2$$

ksaimouli · Answer

so I found fx and fy to be the same m sin(2mx +2ny) and then found fxx (above)

ksaimouli · Answer

now fyy= same as fxx right?

anonymous · Answer

Well, it's similar by symmetry.

ksaimouli · Answer

okay i am stuck at fxy

anonymous · Answer

do x, then do y

ksaimouli · Answer

hold on the Q is $$f_x=m \sin(2mx+2ny)$$
so 
$$f_xx= m \cos(2mx+2ny)(2m)+\cos(2mx+2ny)$$

ksaimouli · Answer

right

ksaimouli · Answer

@wio

anonymous · Answer

actually, now that I think about it... you shouldn't have $2mx+2ny$, there is no reason for the $2$.

ksaimouli · Answer

well question is $$\sin^2(mx+ny)$$
$$fx=2(\sin(mx+my))(\cos(mx+my)(m)$$
$$msin(2mx+2my)$$

ksaimouli · Answer

sin2u=sinu cosu

anonymous · Answer

So you used double angle formula?