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b^3

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #15 on: November 17, 2012, 11:50:31 pm »
+9
The Cross Product

Just to add to CTH's post above, determinates can be used to evaluate cross products.
If we have $\underset{\sim}{a}=a_{1}\underset{\sim}{i}+a_{2}\underset{\sim}{j}+a_{3}\underset{\sim}{k}$ and $\underset{\sim}{b}=b_{1}\underset{\sim}{i}+b_{2}\underset{\sim}{j}+b_{3}\underset{\sim}{k}$ then we define the cross product to be
$\underset{\sim}{a}\times\underset{\sim}{b}=\left(||\underset{\sim}{a}||\,||\underset{\sim}{b}||\sin(\theta)\right)\underset{\sim}{n}$ where $\underset{\sim}{n}$ is a normal vector perpendicular to both $\underset{\sim}{a}$ and $\underset{\sim}{b}$.
We can visualise this using the right hand grip rule.

If we have the vectors $\underset{\sim}{i}$ and $\underset{\sim}{j}$, then the vector perpendicular to both, and sweeping from $\underset{\sim}{a}$ to $\underset{\sim}{b}$ will be $\underset{\sim}{k}$.
So that brings us to these results
\begin{alignedat}{1}\underset{\sim}{i} \times \underset{\sim}{j} & =\underset{\sim}{k}
\\ \underset{\sim}{j} \times \underset{\sim}{k} & =\underset{\sim}{i}
\\ \underset{\sim}{k} \times \underset{\sim}{i} & =\underset{\sim}{j}
\end{alignedat}

$\underset{\sim}{i} \times \underset{\sim}{i}=\underset{\sim}{j}\times\underset{\sim}{j}=\underset{\sim}{k} \times \underset{\sim}{k}=0$
Also note that $\underset{\sim}{a}\times\underset{\sim}{b}=-\underset{\sim}{b} \times \underset{\sim}{a}$

So for
$\underset{\sim}{a}=a_{1}\underset{\sim}{i}+a_{2}\underset{\sim}{j}+a_{3}\underset{\sim}{k}$
$\underset{\sim}{b}=b_{1}\underset{\sim}{i}+b_{2}\underset{\sim}{j}+b_{3}\underset{\sim}{k}$
\begin{alignedat}{1}\underset{\sim}{a}\times\underset{\sim}{b} & =\left(a_{1}\underset{\sim}{i}+a_{2}\underset{\sim}{j}+a_{3}\underset{\sim}{k}\right) \times \left(b_{1}\underset{\sim}{i}+b_{2}\underset{\sim}{j}+b_{3}\underset{\sim}{k}\right)
\\ & =a_{1}b_{1}\underset{\sim}{i} \times \underset{\sim}{i}+a_{1}b_{2}\underset{\sim}{i} \times \underset{\sim}{j}+a_{1}b_{3}\underset{\sim}{i} \times \underset{\sim}{k}+a_{2}b_{1}\underset{\sim}{j} \times \underset{\sim}{i}+a_{2}b_{2}\underset{\sim}{j} \times \underset{\sim}{j}+a_{2}b_{3}\underset{\sim}{j} \times \underset{\sim}{k}+a_{3}b_{1}\underset{\sim}{k} \times \underset{\sim}{i}+a_{3}b_{2}\underset{\sim}{k} \times \underset{\sim}{j}+a_{3}b_{3}\underset{\sim}{k} \times \underset{\sim}{k}
\\ & =a_{1}b_{2}\underset{\sim}{k}-a_{1}b_{3}\underset{\sim}{j}-a_{2}b_{1}\underset{\sim}{k}+a_{2}b_{3}\underset{\sim}{i}+a_{3}b_{1}\underset{\sim}{j}-a_{3}b_{2}\underset{\sim}{i}
\\ & =\left(a_{2}b_{3}-a_{3}b_{2}\right)\underset{\sim}{i}-\left(a_{1}b_{3}-a_{3}b_{1}\right)\underset{\sim}{j}+\left(a_{1}b_{2}-a_{2}b_{1}\right)\underset{\sim}{k}
\\ & =\underset{\sim}{i}\begin{vmatrix}a_{2} & a_{3}
\\ b_{2} & b_{3}
\end{vmatrix}-\underset{\sim}{j}\begin{vmatrix}a_{1} & a_{3}
\\ b_{1} & b_{3}
\end{vmatrix}+\underset{\sim}{k}\begin{vmatrix}a_{1} & a_{2}
\\ b_{1} & b_{2}
\end{vmatrix}
\\ & =\begin{vmatrix}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k}
\\ a_{1} & a_{2} & a_{3}
\\ b_{1} & b_{2} & b_{3}
\end{vmatrix}
\end{alignedat}

So this means we can use determinates to evaluate cross products.
E.g. For
\begin{alignedat}{1}\underset{\sim}{a} & =2\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}
\\ \underset{\sim}{b} & =\underset{\sim}{i}-2\underset{\sim} {j}+3\underset{\sim}{k}
\end{alignedat}

\begin{alignedat}{1}\underset{\sim}{a} \times \underset{\sim}{b} & =\begin{vmatrix}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k}
\\ 2 & 1 & -4
\\ 1 & -2 & 3
\end{vmatrix}
\\ & =\underset{\sim}{i}\begin{vmatrix}1 & -4
\\ -2 & 3
\end{vmatrix}-\underset{\sim}{j}\begin{vmatrix}2 & -4
\\ 1 & 3
\end{vmatrix}+\underset{\sim}{k}\begin{vmatrix}2 & 1
\\ 1 & -2
\end{vmatrix}
\\ & =\underset{\sim}{i}\left(3-8\right)-\underset{\sim}{j}\left(6+4\right)+\underset{\sim}{k}\left(-4-1\right)
\\ & =-5\underset{\sim}{i}-10\underset{\sim}{j}-5\underset{\sim}{k}
\end{alignedat}

Now for a few problems.
Find the cross product of the two vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$ for
1. \begin{alignedat}{1}\underset{\sim}{a} & =\underset{\sim}{i}+\underset{\sim}{j}+\underset{\sim}{k}
\\\underset{\sim}{b} & =2\underset{\sim}{i}-\underset{\sim}{j}-\underset{\sim}{k}
\end{alignedat}

Spoiler
Solution: $3\underset{\sim}{j}+3\underset{\sim}{k}$
2. \begin{alignedat}{1}\underset{\sim}{a} & =2\underset{\sim}{i}+4\underset{\sim}{j}+5\underset{\sim}{k}
\\ \underset{\sim}{b} & =\underset{\sim}{i}+5\underset{\sim}{j}+2\underset{\sim}{k}
\end{alignedat}

Spoiler
Solution: $-17\underset{\sim}{i}+\underset{\sim}{j}+6\underset{\sim}{k}$
3. \begin{alignedat}{1}\underset{\sim}{a} & =-\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}
\\ \underset{\sim}{b} & =3\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}
\end{alignedat}

Spoiler
Solution: $-2\underset{\sim}{j}-2\underset{\sim}{k}$

Now we can also use the cross product to determine the area of a triangle or parallelogram. If we draw the vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$ from the same point, they will form a triangle. Now if we have the angle between these two vectors to be $\theta$, then the height of that triangle will be given by $h=||\underset{\sim}{b}||\sin(\theta)$. Now since we know that the area of a triangle is given by $A=\frac{1}{2}bh$, and the base length will be the magnitude of $\underset{\sim}{a}$, we get
\begin{alignedat}{1}Area_{triangle} & =\frac{1}{2} \times base\times height
\\ & =\frac{1}{2}||\underset{\sim}{a}|| \times ||\underset{\sim}{b}||
\\ & =\frac{1}{2}||\underset{\sim}{a} \times \underset{\sim}{b}||
\end{alignedat}

And the area of the parallelogram will be twice that.
$Area_{parallelogram}=||\underset{\sim}{a} \times \underset{\sim}{b}||$

So lets try a few problems.
1. Find the area of the triangle formed by the vectors \begin{alignedat}{1}\underset{\sim}{a} & =\underset{\sim}{i}+2\underset{\sim}{j}-2\underset{\sim}{k}
\\ \underset{\sim}{b} & =\underset{\sim}{i}+2\underset{\sim}{j}+\underset{\sim}{k}
\end{alignedat}

Spoiler
Ans: $\frac{3\sqrt{5}}{2}\: u^{2}$
2. Find the area of the parallelogram formed by the vectors \begin{alignedat}{1}\underset{\sim}{a} & =-\underset{\sim}{i}+4\underset{\sim}{j}+\underset{\sim}{k}
\\ \underset{\sim}{b} & =3\underset{\sim}{i}+\underset{\sim}{j}+2\underset{\sim}{k}
\end{alignedat}

Spoiler
Ans: $9\sqrt{3}\: u^{2}$

EDIT: Image courtesy of Pi
« Last Edit: February 15, 2013, 09:41:27 pm by b^3 »
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TrueTears

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #16 on: November 24, 2012, 11:30:48 pm »
+1
I just also wanna mention that the cross product is only defined for vectors that are in R^3.

Also:

Consider three vectors a, b, c in R^3

define the scalar triple product to be a. (b x c)

then the volume of a parallelepiped determined by a, b, c is given by |a.(bxc)|
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rife168

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #17 on: November 24, 2012, 11:56:55 pm »
0
I just also wanna mention that the cross product is only defined for vectors that are in R^3.

Is there something analogous and general defined for $\mathbb{R}^n$? That is, given n-1 linearly independent vectors, can you easily compute another vector that is perpendicular to all of those vectors? Or would this simply comprise of tediously finding increasingly more complex formulae from first principles like the following:
$
a_{1}b_{2}\underset{\sim}{k}-a_{1}b_{3}\underset{\sim}{j}-a_{2}b_{1}\underset{\sim}{k}+a_{2}b_{3}\underset{\sim}{i}+a_{3}b_{1}\underset{\sim}{j}-a_{3}b_{2}\underset{\sim}{i}
\\ & =\left(a_{2}b_{3}-a_{3}b_{2}\right)\underset{\sim}{i}-\left(a_{1}b_{3}-a_{3}b_{1}\right)\underset{\sim}{j}+\left(a_{1}b_{2}-a_{2}b_{1}\right)\underset{\sim}{k}
$

Quote
Consider three vectors a, b, c in R^3

define the scalar triple product to be a. (b x c)

then the volume of a parallelepiped determined by a, b, c is given by |a.(bxc)|

I think this is cool, I can work it out by drawing a diagram, but can you share a proper proof/derivation of it?

EDIT:
Just found this:

« Last Edit: November 24, 2012, 11:58:30 pm by rife168 »
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TrueTears

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #18 on: November 25, 2012, 12:03:26 am »
0
yeah ok it's not that hard, im sure you can see it from the diagram, that is how i derive it

V = Ah = |axb| |c||cos(phi)| = |c.(axb)|

also the scalar triple product is a cool way of showing that 3 vectors are coplanar. ie a.(bxc) = 0 then a, b, c are coplanar (why?)
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rife168

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #19 on: November 25, 2012, 12:22:32 am »
+1
yeah ok it's not that hard, im sure you can see it from the diagram, that is how i derive it

V = Ah = |axb| |c||cos(phi)| = |c.(axb)|

also the scalar triple product is a cool way of showing that 3 vectors are coplanar. ie a.(bxc) = 0 then a, b, c are coplanar (why?)

Yep I found that diagram after I posted, it makes it very clear :]

Because if three vectors are coplanar, then the volume of the paralellepiped determined by the vectors will lie in a plane, i.e. 2 dimensional, i.e. volume=a.(bxc)=0
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Hancock

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #20 on: November 25, 2012, 02:21:42 am »
+2
In relation to a more generalized way of finding a vector, v, such that it is perpendicular to all the vectors in the set S (which holds n-1 vectors, and we are in R^n for the moment), you can use the Gram-Schmidt Process.

You can first choose an arbitrary vector, w, to be our new vector in the set which we will change into a perpendicular vector. The GS Process works in a way such that it doesn't matter which vector w we choose, because it will change it into the correct perpendicular vector (the process also normalizes it, such that it has a length of 1).

Let's do an example.
Let's say we have 2 orthogonal vectors. We can use this process to make any set of vectors orthogonal to each other, but at this time, we are just trying to find a vector that is orthogonal to a set of already orthogonal vectors.

z = (1/3)*(2,2,1)
v = (1,-1,0)

Notice that z.v = 0

So, we want to find a vector, w such that z.w = 0 and v.w = 0.
Let's just choose w = (1,1,1)

In order to find this orthogonal vector, we use the GS procedure. Let u3 = orthogonal vector to z and v.

u3 = w - <z,w>z - <v,w>v

Where <x,y> = x.y (in this case).

<z,w> = (1,1,1) dot (1/3)(2,2,1) = (1/3)(2+2+1) = 5/3
<v,w> = (1,0,0) dot (1,1,1) = 0

Therefore:

u3 = (1,1,1) - (5/3)*(1/3)*(2,2,1) - 0*(1,0,0)
u3 = (-1/9,-1/9,4/9)

So, we have found our orthogonal vector to z and v using this process. This can be generalized to a set of k vectors, by assuming that we have k vectors (v1,v2,v3...vk)

uk = vk - <v1,vk>*v2 - <v2,vk>*v2 ..... - <v_k-1, vk>*v_k-1

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #21 on: January 13, 2013, 12:10:01 am »
+4
am i the only one here that doesn't understand anything going on here  ? LOL.

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #22 on: January 13, 2013, 01:12:01 pm »
0
Lots of spesh stuff I guess
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Re: BORED already? Want to learn a bit of maths? :P
« Reply #23 on: February 15, 2013, 09:08:46 pm »
0
b^3, thanks a heap for fixing up the latex!
much appreciated and very helpful for my work
« Last Edit: February 15, 2013, 09:41:56 pm by Machi »
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b^3

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #24 on: February 15, 2013, 09:42:31 pm »
+1
Forgot to fix my last post when I came back and fixed the $\LaTeX$ after the system changed. Should be all good now.
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Jenny_2108

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #25 on: March 04, 2013, 02:05:39 am »
0
$\det\left(\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\right )=\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=a\begin{vmatrix}e&f\\h&i\end{vmatrix}-b\begin{vmatrix}d&f\\g&i\end{vmatrix}+c\begin{vmatrix}d&e\\g&h\end{vmatrix}$

(note: the use of straight lines on the matrices is a shorthand for the determinant -- you may not have seen it before, I hadn't seen it before uni...)

So we're left with a few (2 x 2) determinants, which we know how to find. But what exactly did I do there? Well, the $a, b$, and $c$ came from the first row. The $\begin{vmatrix}e&f\\h&i\end{vmatrix}$ matrix determinant came from temporarily 'deleting' the row and column that $a$ came from:

$\begin{vmatrix}(a)&(b)&(c)\\(d)&e&f\\(g)&h&i\end{vmatrix}\rightarrow \begin{vmatrix}e&f\\h&i\end{vmatrix}
$

And similarly, the matrix determinant that $b$ is multiplied by came from 'deleting' the row and column that $b$ is in:

$\begin{vmatrix}(a)&(b)&(c)\\d&(e)&f\\g&(h)&i\end{vmatrix}\rightarrow \begin{vmatrix}d&f\\g&i\end{vmatrix}$

But there's one more thing: the negative sign in front of the $b$. Whether the coefficient of each element is 1 or -1 can be determined from the following representation:

$\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}$

So basically, starting at the first element (first row, first column), it's positive, then it alternates between positive and negative. This can be extended to higher dimensions, e.g. for a (6 x 6):

$\begin{bmatrix}+&-&+&-&+&-\\-&+&-&+&-&+\\+&-&+&-&+&-\\-&+&-&+&-&+\\+&-&+&-&+&-\\-&+&-&+&-&+\end{bmatrix}$

So, for example, if we expanded the second row instead, we would have:

$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=-d\begin{vmatrix}b&c\\h&i\end{vmatrix}+e\begin{vmatrix}a&c\\g&i\end{vmatrix}-f\begin{vmatrix}a&b\\g&h\end{vmatrix}$

If you expand this out you'll find it's exactly the same as the other expression.

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #26 on: July 16, 2013, 02:00:56 am »
+6
I came across this textbook http://maths.anu.edu.au/files/introduction_contemporary_mathematics.pdf that's targeted at Year 11/Year 12 students in the ACT and figured a few people here will find it interesting.

Seems like it's designed to accompany this (2 year long?) course for high school students http://maths.anu.edu.au/education/degree-programs/anu-secondary-college and looks like it references The Heart of Mathematics textbook a bit too (I don't really know what that course or textbook is like).

It discusses some number theory, irrational/real number stuff, a chapter about infinity, fractals and geometry/topology. There's more in-depth table of contents at the start of each chapter.
« Last Edit: July 16, 2013, 02:04:26 am by Lazyred »

b^3

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Re: BORED already? Want to learn a bit of maths? :P
« Reply #27 on: November 23, 2013, 03:45:07 pm »
+4
Typed this up a while ago, forgot to post it up.

Partial Derivatives

If we have a function of two variables, $f\left(x,y\right)$, then we can take partial derivatives of that function. To do this we differentiate with respect to one variable, and hold all others constant. So that is if you had a surface, and went to find the partial derivative with respect to $x$, $\frac{\partial f}{\partial x}$ at a point. This would be the same as drawing a tangent line at the point on the surface that goes along the constant $y$ value for that point.

http://mathinsight.org/site_media/image/image/partial_derivative_as_slope.png

Now the partial derivative of $f$ with respect to $x$ can be denoted by $\frac{\partial f}{\partial x}$ or $f_{x}$.

So if we have $f\left(x,y\right)=x^{2}y-2xy+x$, then to find $\frac{\partial f}{\partial x}$ we would look at $y$ as any other constant. So for the first term we would think of it as differentiating $kx^{2}$ with respect to $x$ to give $2kx$. So we have
\begin{alignedat}{1}\frac{\partial f}{\partial x} & =2xy-2y+1\end{alignedat}

If we wanted to find the partial derivative of $f$ with respect to $y$ we would hold $x$ constant. So for the first term we are differentiating $ky$ with respect to $y$. Notice that the last term drops off as we are effectively differentiating a constant to give $0$.
\begin{alignedat}{1}\frac{\partial f}{\partial y} & =x^{2}-2x\end{alignedat}

Now we can also go one step further and take another partial derivative of the above, in this case we represent the second partial derivative of $f$ with respect to $x$ as $\frac{\partial^{2}f}{\partial x^{2}}$ or $f_{xx}$, which is the same as $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)$. So from the above we have
\begin{alignedat}{1}\frac{\partial^{2}f}{\partial x^{2}} & =\frac{\partial}{\partial x}\left(2xy-2y+1\right)
\\ & =2y
\end{alignedat}

The same goes for $f_{yy}$
\begin{alignedat}{1}\frac{\partial^{2}f}{\partial y^{2}} & =0\end{alignedat}

Now we can also take the partial derivative of $f_{x}$ with respect to $y$, that is we have a mixed partial derivative. The derivative that you do first is on the right of the fraction notation, $\frac{\partial^{2}f}{\partial y\partial x}$ and on the left of the subscript notation, $f_{xy}$. So from the above we have
\begin{alignedat}{1}\frac{\partial^{2}f}{\partial y\partial x} & =2x-2
\\ \frac{\partial^{2}f}{\partial x\partial y} & =2x-2
\end{alignedat}

Now what do you notice about $f_{xy}$ and $f_{yx}$? They are the same, now this is a result of Clairaut's Theorem which stats that if a function $f$ is defined on $D$ and has continuous partial derivatives $f_{x}$ and $f_{y}$ on $D$ then the mixed partial derivatives are equal, that is
\begin{alignedat}{1}f_{xy} & =f_{yx}\end{alignedat}

Now lets try a few examples.

1. Find the first, second and mixed partials derivaties of $f$ with respect to $x$ and $y$. That is find $f_{x}$, $f_{y}$, $f_{xx}$, $f_{yy}$ and $f_{xy}$. (Note: In this case there is no need to find $f_{yx}$.

a) $f\left(x,y\right)=x^{3}+x^{2}y^{3}-4x+8y+2$
Spoiler
\begin{alignedat}{1}f_{x} & =3x^{2}+2xy^{3}-4
\\ f_{y} & =3x^{2}y^{2}+8
\\ f_{xx} & =6x+2y^{3}
\\ f_{yy} & =6x^{2}y
\\ f_{xy} & =6xy^{2}
\end{alignedat}
b) $f\left(x,y\right)=\cos\left(xy^{2}\right)+\sin\left(x+y\right)$
Spoiler
\begin{alignedat}{1}f_{x} & =-y^{2}\sin\left(xy^{2}\right)+\cos\left(x+y\right)
\\ f_{y} & =-2xy\sin\left(xy^{2}\right)+\cos\left(x+y\right)
\\ f_{xx} & =-y^{4}\cos\left(xy^{2}\right)-\sin\left(x+y\right)
\\ f_{yy} & =\left(-2xy\right)\left(2xy\cos\left(xy^{2}\right)\right)+\left(-2x\right)\sin\left(xy^{2}\right)-\sin\left(x+y\right)
\\ & =-4x^{2}y^{2}\cos\left(xy^{2}\right)-2x\sin\left(xy^{2}\right)-\sin\left(x+y\right)
\\ f_{xy} & =-y^{2}\left(2xy\cos\left(xy^{2}\right)\right)+\sin\left(xy^{2}\right)\left(-2y\right)-\sin\left(x+y\right)
\\ & =-2xy^{3}\cos\left(xy^{2}\right)-2y\sin\left(xy^{2}\right)-\sin\left(x+y\right)
\end{alignedat}
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